Make 0 0 0 0 = 8
57
$begingroup$
Can you find a way to make:
$0 0 0 0 = 8$
by adding any operations or symbols? You can use only these symbols:
$+, -, *, !, /, hat, , ()$.
It is limited to this list, and concatenation is also allowed. You cannot add other numbers to the equation.
mathematics number-theory
asked Sep 7 '18 at 0:35
kraby15kraby15
2,3943930
$endgroup$
This question has an open bounty worth +100
reputation from user477343 ending in 7 days.
One or more of the answers is exemplary and worthy of an additional bounty.
This bounty will be awarded to @let_the_coding_begin's very first and very elegant answer — it has over 100 upvotes (including my own)! In my opinion, it is just as good the accepted answer, if not, better. So well cone, and congratulations! :D
add a comment |
57
$begingroup$
Can you find a way to make:
$0 0 0 0 = 8$
by adding any operations or symbols? You can use only these symbols:
$+, -, *, !, /, hat, , ()$.
It is limited to this list, and concatenation is also allowed. You cannot add other numbers to the equation.
mathematics number-theory
asked Sep 7 '18 at 0:35
kraby15kraby15
2,3943930
$endgroup$
This question has an open bounty worth +100
reputation from user477343 ending in 7 days.
One or more of the answers is exemplary and worthy of an additional bounty.
This bounty will be awarded to @let_the_coding_begin's very first and very elegant answer — it has over 100 upvotes (including my own)! In my opinion, it is just as good the accepted answer, if not, better. So well cone, and congratulations! :D
40
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Just put a slash over the equal sign!
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– Yout Ried
Sep 7 '18 at 0:40
13
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@YoutRied: standard loophole: neither funny nor creative. It's usually the first answer on any math puzzle.
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– Thomas Weller
Sep 7 '18 at 18:40
2
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Are we allowed to move anything, or is it insert only?
$endgroup$
– ctrl-alt-delor
Sep 7 '18 at 19:33
2
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Are you allowed to concatenate operators? (e.g!!,**,--, and++)
$endgroup$
– Ole Tange
Sep 9 '18 at 1:09
1
$begingroup$
@OleTange The accepted answer does concatenate operators. Also, the question allows it.
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– haykam
Sep 9 '18 at 13:59
add a comment |
57
57
57
7
$begingroup$
Can you find a way to make:
$0 0 0 0 = 8$
by adding any operations or symbols? You can use only these symbols:
$+, -, *, !, /, hat, , ()$.
It is limited to this list, and concatenation is also allowed. You cannot add other numbers to the equation.
mathematics number-theory
asked Sep 7 '18 at 0:35
kraby15kraby15
2,3943930
$endgroup$
Can you find a way to make:
$0 0 0 0 = 8$
by adding any operations or symbols? You can use only these symbols:
$+, -, *, !, /, hat, , ()$.
It is limited to this list, and concatenation is also allowed. You cannot add other numbers to the equation.
mathematics number-theory
mathematics number-theory
asked Sep 7 '18 at 0:35
kraby15kraby15
2,3943930
asked Sep 7 '18 at 0:35
kraby15kraby15
2,3943930
asked Sep 7 '18 at 0:35
kraby15kraby15
2,3943930
asked Sep 7 '18 at 0:35
kraby15kraby15
2,3943930
asked Sep 7 '18 at 0:35
kraby15kraby15
2,3943930
2,3943930
This question has an open bounty worth +100
reputation from user477343 ending in 7 days.
One or more of the answers is exemplary and worthy of an additional bounty.
This bounty will be awarded to @let_the_coding_begin's very first and very elegant answer — it has over 100 upvotes (including my own)! In my opinion, it is just as good the accepted answer, if not, better. So well cone, and congratulations! :D
This question has an open bounty worth +100
reputation from user477343 ending in 7 days.
One or more of the answers is exemplary and worthy of an additional bounty.
This bounty will be awarded to @let_the_coding_begin's very first and very elegant answer — it has over 100 upvotes (including my own)! In my opinion, it is just as good the accepted answer, if not, better. So well cone, and congratulations! :D
40
$begingroup$
Just put a slash over the equal sign!
$endgroup$
– Yout Ried
Sep 7 '18 at 0:40
13
$begingroup$
@YoutRied: standard loophole: neither funny nor creative. It's usually the first answer on any math puzzle.
$endgroup$
– Thomas Weller
Sep 7 '18 at 18:40
2
$begingroup$
Are we allowed to move anything, or is it insert only?
$endgroup$
– ctrl-alt-delor
Sep 7 '18 at 19:33
2
$begingroup$
Are you allowed to concatenate operators? (e.g!!,**,--, and++)
$endgroup$
– Ole Tange
Sep 9 '18 at 1:09
1
$begingroup$
@OleTange The accepted answer does concatenate operators. Also, the question allows it.
$endgroup$
– haykam
Sep 9 '18 at 13:59
add a comment |
40
$begingroup$
Just put a slash over the equal sign!
$endgroup$
– Yout Ried
Sep 7 '18 at 0:40
13
$begingroup$
@YoutRied: standard loophole: neither funny nor creative. It's usually the first answer on any math puzzle.
$endgroup$
– Thomas Weller
Sep 7 '18 at 18:40
2
$begingroup$
Are we allowed to move anything, or is it insert only?
$endgroup$
– ctrl-alt-delor
Sep 7 '18 at 19:33
2
$begingroup$
Are you allowed to concatenate operators? (e.g!!,**,--, and++)
$endgroup$
– Ole Tange
Sep 9 '18 at 1:09
1
$begingroup$
@OleTange The accepted answer does concatenate operators. Also, the question allows it.
$endgroup$
– haykam
Sep 9 '18 at 13:59
40
40
$begingroup$
Just put a slash over the equal sign!
$endgroup$
– Yout Ried
Sep 7 '18 at 0:40
$begingroup$
Just put a slash over the equal sign!
$endgroup$
– Yout Ried
Sep 7 '18 at 0:40
13
13
$begingroup$
@YoutRied: standard loophole: neither funny nor creative. It's usually the first answer on any math puzzle.
$endgroup$
– Thomas Weller
Sep 7 '18 at 18:40
$begingroup$
@YoutRied: standard loophole: neither funny nor creative. It's usually the first answer on any math puzzle.
$endgroup$
– Thomas Weller
Sep 7 '18 at 18:40
2
2
$begingroup$
Are we allowed to move anything, or is it insert only?
$endgroup$
– ctrl-alt-delor
Sep 7 '18 at 19:33
$begingroup$
Are we allowed to move anything, or is it insert only?
$endgroup$
– ctrl-alt-delor
Sep 7 '18 at 19:33
2
2
$begingroup$
Are you allowed to concatenate operators? (e.g
!!, **, --, and ++)$endgroup$
– Ole Tange
Sep 9 '18 at 1:09
$begingroup$
Are you allowed to concatenate operators? (e.g
!!, **, --, and ++)$endgroup$
– Ole Tange
Sep 9 '18 at 1:09
1
1
$begingroup$
@OleTange The accepted answer does concatenate operators. Also, the question allows it.
$endgroup$
– haykam
Sep 9 '18 at 13:59
$begingroup$
@OleTange The accepted answer does concatenate operators. Also, the question allows it.
$endgroup$
– haykam
Sep 9 '18 at 13:59
add a comment |
23 Answers
23
active
oldest
votes
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I think that
$left( 0! + 0! + 0! + 0! right)!! = 8$.
This is because
$0! = 1$ and $4!! = 8$. Note that $left( 0! + 0! + 0! + 0! right)!! = left( 1+1+1+1 right)!! = left (4 right)!! = 8$.
This works and is valid because
The question says I’m allowed to use any of the following symbols in my answer, I am not restricted to using $!$ as an operation only.
edited Sep 7 '18 at 6:42
Laurel
818310
answered Sep 7 '18 at 0:40
El-GuestEl-Guest
21.6k25092
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65
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for those who are confused, !! is a mathematical operator that gives the product of all positive integers upto the argument which have the same parity (odd/even) as the argument, its called a semifactorial. (thus 4!! is 2*4 = 8)
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– casualcoder
Sep 7 '18 at 7:16
13
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@casualcoder most people assume $n!!$ for an arbitrary non-negative integer $n$ is the same as $(n!)!$, but it's not. That brings up some confusion, so I usually write $n!_k$ to denote $n$ with $k$ factorials, but not on this site.
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– user477343
Sep 7 '18 at 11:49
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@casualcoder Google disagrees with Wolfram on this.
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– user1717828
Sep 7 '18 at 14:46
10
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Google disagrees with Wolfram on a lot of things. When it comes to math, chances are Wolfram is the correct one.
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– AlexanderJ93
Sep 7 '18 at 23:03
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@user1717828, user477343’s comment above explains why Google’s interpretation is not correct in this context.
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– El-Guest
Sep 7 '18 at 23:09
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A lateral thinking answer:
0! 0 0 0, because the binary equivalent of 8 is 1000 :)
edited Sep 7 '18 at 14:01
El-Guest
21.6k25092
answered Sep 7 '18 at 13:44
let_the_coding_beginlet_the_coding_begin
1,051114
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9
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I like this! very direct and minimal.
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– Ruadhan2300
Sep 7 '18 at 14:25
2
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My favourite one! I did wonder if someone would go binary.
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– oliver-clare
Sep 7 '18 at 15:07
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I was just checking existing answers to see if anyone else had thought of ! In the sense of negation, and binary. So "yes" :)
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– Stilez
Sep 10 '18 at 8:24
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I have reached my daily voting limit (DVL), but Imma vote this when I can, get it to $99$, and then hope that another user actually puts another upvote :P
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– user477343
Oct 3 '18 at 7:11
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68
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$0 + 0 + 0 + 0 ~~!!=~ 8$
because
$ !!= $ is an alternative way of writing $ ne $.
answered Sep 7 '18 at 8:43
Teemu PiippoTeemu Piippo
66724
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4
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This is the answer!
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– user51438
Sep 8 '18 at 2:20
3
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@TheodoreWeld nope, this question is tagged mathematics, and you have to use a slash similar to Yout Ried instead of this
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– phuclv
Sep 8 '18 at 4:24
add a comment |
55
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Lateral thinking!
$$0+0+substack{0\0}=0+0+8=8$$
answered Sep 7 '18 at 6:25
TheSimpliFireTheSimpliFire
2,155532
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8
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First I thought "how does picking 0 out of 0 give you 8 combinations", then I got it.
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– Fabian Röling
Sep 7 '18 at 12:01
82
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Looks more like vertical thinking to me.
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– Evargalo
Sep 7 '18 at 12:02
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34
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let me try:
$0! Vert 0 - 0!-0! =8$
$10-1-1=8$
$Vert$ is a concatenation operation
answered Sep 7 '18 at 0:43
malioboromalioboro
2,77311033
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1
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good solution but in this case you cant use '[' or ']'
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– casualcoder
Sep 7 '18 at 7:13
3
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@casualcoder It was explicitly allowed. If this wasn't meant to be a valid solution, why even allow it, as 00 = 0+0 = 0?
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– Zizy Archer
Sep 7 '18 at 7:51
1
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I've never used $[ldots ]$ as an operator before, let alone for what you use it as in this answer. I usually use it as a set notation; i.e., $[n]={1,2,ldots n}$ :P
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– user477343
Sep 7 '18 at 11:53
2
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I think "concatenation allowed" in these puzzles usually means that, for example,2 8 7can be28 ÷ 7 = 4, but not(2+8)7 = 107
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– Chronocidal
Sep 7 '18 at 14:22
3
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@Chronocidal What you are saying is probably true about most of these puzzles. But in this particular puzzle where all the digits are 0, that interpretation does not make a lot of sense. To me this answer feels more legitimate than the accepted answer. That's because concatenation is explicitly allowed, but the !! is not explicitly allowed (though it may be implied).
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– kasperd
Sep 8 '18 at 12:20
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show 3 more comments
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$((0!+0!)^{(0!+0!)})!!$
Evaluation:
$((0!+0!)^{(0!+0!)})!!$
$rightarrow ((1+1)^{(1+1)})!!$
$rightarrow (2^2)!!$
$rightarrow 4!! = 8$
answered Sep 7 '18 at 14:04
u_ndefinedu_ndefined
2,7801437
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How did I not think of that?? DVL16 :
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– user477343
Oct 3 '18 at 7:12
add a comment |
27
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It's different:
$,++$
$0;;;0$
$,++$
$0;;;0$
$,++$
An ASCII art $8$ using only four $0$'s and $+$'s.
answered Sep 7 '18 at 14:46
JonMark PerryJonMark Perry
20.6k64099
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add a comment |
16
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0 + 0 + 0 + 0 = !8
because
In C/C++, ! refers to the logical not operator, where all non-zero values become 0, and 0 becomes 1.
answered Sep 7 '18 at 22:48
l kl k
1612
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I think this should be "the binary not operator".
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– Raimund Krämer
Sep 11 '18 at 8:17
3
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@RaimundKrämer ITYM unary not. C99 calls it the "logical negation operator" and files it in the chapter "Unary operators".
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– Jens
Sep 11 '18 at 15:56
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But !8 can be a subfactorial of 8.
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– rus9384
Sep 12 '18 at 11:39
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@Raimund Krämer, In programming, we have logical operators and bitwise operators. Logical operators (!,&&,||in C) operate on boolean values (true and false). Bitwise operators (~,&,|,^in C) operate on bits (0 and 1) of numbers. Calling either set binary operators would be confusing, as both set work on binary values.
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– ikegami
Sep 13 '18 at 11:12
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@Raimund Krämer Also, "binary operator" can mean any operator on 2 values (as opposed to a unary operator such as ! or ~)
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– l k
Sep 18 '18 at 6:23
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It's just a matter of perspective ...
0!/0 + 0!/0 = ∞
My reasoning....
0/0 is undefined so we have to first change the 0's into 1's with 0!
(...and why did you write the infinity symbol sideways in your question?)
answered Sep 7 '18 at 18:20
rrauenzarrauenza
24915
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1
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x/0 isn't infinite, though...
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– Adam Smith
Sep 8 '18 at 17:49
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Lateral thinking was yesterday. Vertical thinking is the new kid in town!
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– Jens
Sep 11 '18 at 15:59
1
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x/0 == +infinity, per ieee 754
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– j__m
Sep 15 '18 at 12:40
add a comment |
10
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$$[+!0]+[0]-!0-!0$$
Works in JavaScript. Hit F12 and type this into the console (This equation editor uses "−" instead of "-" so copy and paste doesn't quite work). Otherwise, it works in the same way as @malioboro and @Arnaldur's answers.
In fact, you can make any JavaScript application run just by using a combination of 6 characters, which is what inspired me to make this. I substituted + for 0 when asking JSF**k to do 10-2.
answered Sep 7 '18 at 12:35
NamytsNamyts
4647
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I like this abuse of javascript a lot. !0 for true, +!0 for 1, [1] for array containing 1, [1]+[0] for concatenate two arrays to get the string "10" ; subtract 1 from that twice to end up as 8. Love it.
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– Ross Presser
Sep 8 '18 at 4:10
2
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but the question is tagged mathematics and not programming
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– phuclv
Sep 8 '18 at 4:21
3
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Who says you can use square brackets....
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– user52269
Sep 8 '18 at 6:57
add a comment |
8
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$concat(0!,0) - 0! - 0! = 8$
becomes:
$concat(1,0) - 1 - 1 = 8$
and finally:
$10 - 2 = 8$
cool puzzle!
answered Sep 7 '18 at 5:38
ArnaldurArnaldur
971
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2
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Great answer! But it has already been answered (check out @malioboro 's answer). Nonetheless, welcome to the Puzzling Stack Exchange (Puzzling.SE)! Since you are new to the Stack Exchange community, let alone this site, I strongly suggest that you visit the Help Center for more info; in particular, I suggest going here, then here for questions (not answers) :D
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– user477343
Sep 7 '18 at 7:44
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7
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$0 + 0 + 0 + 0 equiv 8$
Adding the symbol $-$ over the equals sign makes it a congruence sign.
Considering the congruence relation, we must be working mod N, where N divides 8.
answered Sep 8 '18 at 6:39
VaelusVaelus
40124
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3
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Just put the minus symbol over the first zero to give it the look of a figure 8 and use plus to add the zeros.
answered Sep 7 '18 at 11:37
SentinelSentinel
1,041212
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Great answer, but unfortunately, this is a duplicate of @TheSimpliFire 's answer.
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– user477343
Sep 7 '18 at 11:51
2
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@user477343 No, not quite, since this answer puts a line through the 0 instead of putting two 0s on top of each other. In this answer, only one 0 becomes an 8.
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– Fabian Röling
Sep 7 '18 at 12:02
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@FabianRöling As I read it, it uses the minus symbol to be put over the first zero and make it look like an $8$. In my imagination, I assumed the zero is the bottom circle of $8$ since there is something over it, of which is another circle $largecirc$. I did not fully understand what this answer was talking about in saying to use a subtraction symbol, but just assumed it was just like TheSimpliFire's answer due to their similarity; and since that is a great answer, I thought this was great, too. If you understand this answer, may you please explain to me in other words if you can? :
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– user477343
Sep 7 '18 at 12:09
3
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@user477343 It uses the horizontal line of the minus sign to cut through the middle of the $0$. It is "over" the zero in the sense of overwritten.
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– Jaap Scherphuis
Sep 7 '18 at 12:55
1
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Yeah. 'over' in the sense of "in front of" on a dimension between you and your monitor.
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– Sentinel
Sep 7 '18 at 14:44
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3
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Here is an answer that doesn't use the semi-factorial or any concatenation.
$$0 + 0 - 0! / 0 = (-8)!$$
The left side is $-1/0$ and the right side is $-infty$.
Plugging the expression into Wolfram.
answered Sep 8 '18 at 0:20
user1717828user1717828
1,514617
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4
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Division by zero is undefined (the limit of 0!/x as x goes to 0 is plus or minus infinity) and so are negative integer factorials (they are not equivalent to negative infinity). Even if they both evaluated to +∞, you can prove nonsense assuming you can equate infinities and treat like a real number. E.g., |1/0| = +∞, adding 1 to positive infinity is still positive infinity, hence |1/0| + 1 = +∞, hence |1/0| = |1/0| +1 by transitivity of equality. Subtract |1/0| from both sides and prove 0 = 1.
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– dr jimbob
Sep 8 '18 at 18:11
add a comment |
3
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concatenate(0!, 0) - concatenate(0! + 0!) = 8.
Note that 0! = 1
(0!, 0) = 10, and (0! + 0!) = 2, so 10 - 2 = 8
edited Sep 11 '18 at 1:59
Alto
1,222320
answered Sep 11 '18 at 0:57
NickNick
291
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Unfortunately concatenate is not a valid operation.
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– boboquack
Sep 11 '18 at 1:37
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Congrats, concatenation is allowed now! Good answer, but people beat you to it. Welcome to Puzzling.SE! Here's a bonus puzzle: What's the word? Hilarious - extremely amusing
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– Alto
Sep 11 '18 at 1:47
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Also, concatenation is combination. Basically, (2, 4) = 24, you get it.
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– Alto
Sep 11 '18 at 1:48
add a comment |
3
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This could work too:
(0!+0+0)/0 = ∞
Explanation
(0!+0+0)/0 = 1/0 which is infinity (8 but put horizontally)
answered Sep 13 '18 at 9:10
Michal B.Michal B.
2665
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That is not infinity — it is undefined :
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– user477343
Oct 3 '18 at 11:33
add a comment |
3
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Add a - above the equals to get $0000 equiv 8$, which is true assuming we are working in the ring $mathbb{Z}/mathbb{2Z}$. (Note I'm trying to avoid writing $[0] = [8]$...)
edited Sep 14 '18 at 2:45
a stone arachnid
1697
answered Sep 11 '18 at 15:52
Gus314Gus314
312
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add a comment |
2
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Question limits the symbols, not the operations. So with the symbol + can make the operator ++.
(++(++(++(++(++(++(++0000))))))) = 8
answered Sep 7 '18 at 21:23
David Browne - MicrosoftDavid Browne - Microsoft
1294
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I think you'd need the prefix form if you actually wanted that to work.
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– LegionMammal978
Sep 7 '18 at 23:48
1
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Doh. Of course you can't increment a literal in the first place.
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– David Browne - Microsoft
Sep 7 '18 at 23:55
1
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but that symbol doesn't exist in mathematics
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– phuclv
Sep 8 '18 at 4:20
2
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You need to pass an l-value to the++and--operators, and neither their result or a literal is one.
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– NieDzejkob
Sep 8 '18 at 14:35
add a comment |
0
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$ ((0! + 0!)$)*(0! + 0!) = 2^2*2 = 8 $.
Further explanation:
The $$$ operation denotes the superfactorial defined as : $ n$ = (n!) uparrow uparrow (n!)$.
answered Sep 7 '18 at 14:28
RiaRia
354111
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3
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Good idea, but make sure to read the question fully. It gives you a list of what operations can be used and $ is not one of them.
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– Sensoray
Sep 7 '18 at 15:26
add a comment |
0
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If you turn the problem around
answered Sep 8 '18 at 8:09
Florian FFlorian F
9,22612260
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Similar to @rrauenza's answer
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– TheSimpliFire
Sep 8 '18 at 8:11
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Yes. A different way to express the same idea.
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– Florian F
Sep 8 '18 at 9:13
add a comment |
0
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Similar to @Vaelus
$0+0+0+0 leq 8$
Explanation
You can get the $leq$ by adding a $-$ inclined on top of the $=$
answered Sep 8 '18 at 19:44
villasvvillasv
1487
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The goal is to strictly make $0,0,0,0=8$ and not to "make the statement true" (therefore allowing this answer to be acceptable) :
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– user477343
Oct 3 '18 at 11:35
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When you say "the goal is to make [math expression]", then you're under the influence of the interpretation of that mathematical expression. IMO, "make the statement true" is the one and only interpretation to be taken, while you could argue more strict rules about not manipulating already existing symbols.
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– villasv
Oct 3 '18 at 13:18
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I see, you have a point. I should've said that my comment is solely based on how I interpreted the question, possibly explaining why your answer was sadly downvoted. I am positive that there exists another answer that manipulates the equation in a similar way that you have. Sorry if I sounded mean. You can get a $(+1)$ but in at least $2$ hours, once my daily voting limit (DVL) is over :)
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– user477343
Oct 3 '18 at 21:00
add a comment |
0
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As per the list of allowed symbols we are clearly allowed to use "$,$" and "$.$"
This is doubly evident as otherwise how would we use the $mathbb{concatenation}$ function without a comma to separate the arguments?
So the solution is:
$0! - mathbb{concatenation}(., 0 + 0! + 0!!) = .8$
$1 - mathbb{concatenation}(., 0 + 1 + 1) = .8$
$1 - mathbb{concatenation}(., 2) = .8$
$1 - .2 = .8$
answered Sep 10 '18 at 16:55
SamYonnouSamYonnou
69067
$endgroup$
$begingroup$
You may simplify0!!as0!.
$endgroup$
– Cœur
Sep 11 '18 at 15:28
$begingroup$
I felt like $0+0!+0!!$ looked more elegant/purposeful as if it was part of a sequence like $a(n, m) = a(n-1,m)!; a(0,m)=m$
$endgroup$
– SamYonnou
Sep 11 '18 at 16:58
add a comment |
-1
$begingroup$
$00^{00} = 0^0$ which is indeterminate. In some sense, an indeterminate form can be equal to any value, since in Calculus, a function that approaches "$0^0$" can approach any real value, including $8$. So in that sense, $0^0 = 8$.
If you don't like concatenating two zeroes as "$00$", then $0^0 + 0 + 0$ also works.
answered Sep 7 '18 at 13:22
RothXRothX
1173
$endgroup$
7
$begingroup$
There are no widely accepted definitions under which your equation is considered to be true.
$endgroup$
– Tanner Swett
Sep 7 '18 at 17:34
$begingroup$
@TannerSwett Didn't think it was that unreasonable. My thinking is you can construct a function $f(x)^{g(x)}$ where $f$ and $g$ both limit to zero at some point $c$ such that $f(x)^{g(x)}$ limits to any real number at $c$, including $8$. Admittedly, writing $0^0 = 8$ is basically mathematically false. It is indeterminate for the reason I stated above. But this is Puzzling SE, not Mathematics SE, so I figured it was ok to think outside the box and not be formal with math.
$endgroup$
– RothX
Sep 10 '18 at 13:23
$begingroup$
Yeah, I dunno if I'd call it "that unreasonable". But if I saw a student say that the equation $0^0 = 8$ is true, I'd think that they probably misunderstand what an indeterminate form means. If I saw a math professor say the same thing, I'd ask them if they could please be a little more rigorous.
$endgroup$
– Tanner Swett
Sep 10 '18 at 13:59
$begingroup$
@TannerSwett Yeah, I completely agree. I'd never write $0^0 = 8$ in a mathematical setting, but since this is a puzzle, not a math problem, I thought maybe I could stretch a bit.
$endgroup$
– RothX
Sep 11 '18 at 1:58
add a comment |
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23 Answers
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23 Answers
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96
$begingroup$
I think that
$left( 0! + 0! + 0! + 0! right)!! = 8$.
This is because
$0! = 1$ and $4!! = 8$. Note that $left( 0! + 0! + 0! + 0! right)!! = left( 1+1+1+1 right)!! = left (4 right)!! = 8$.
This works and is valid because
The question says I’m allowed to use any of the following symbols in my answer, I am not restricted to using $!$ as an operation only.
edited Sep 7 '18 at 6:42
Laurel
818310
answered Sep 7 '18 at 0:40
El-GuestEl-Guest
21.6k25092
$endgroup$
65
$begingroup$
for those who are confused, !! is a mathematical operator that gives the product of all positive integers upto the argument which have the same parity (odd/even) as the argument, its called a semifactorial. (thus 4!! is 2*4 = 8)
$endgroup$
– casualcoder
Sep 7 '18 at 7:16
13
$begingroup$
@casualcoder most people assume $n!!$ for an arbitrary non-negative integer $n$ is the same as $(n!)!$, but it's not. That brings up some confusion, so I usually write $n!_k$ to denote $n$ with $k$ factorials, but not on this site.
$endgroup$
– user477343
Sep 7 '18 at 11:49
3
$begingroup$
@casualcoder Google disagrees with Wolfram on this.
$endgroup$
– user1717828
Sep 7 '18 at 14:46
10
$begingroup$
Google disagrees with Wolfram on a lot of things. When it comes to math, chances are Wolfram is the correct one.
$endgroup$
– AlexanderJ93
Sep 7 '18 at 23:03
2
$begingroup$
@user1717828, user477343’s comment above explains why Google’s interpretation is not correct in this context.
$endgroup$
– El-Guest
Sep 7 '18 at 23:09
|
show 3 more comments
96
$begingroup$
I think that
$left( 0! + 0! + 0! + 0! right)!! = 8$.
This is because
$0! = 1$ and $4!! = 8$. Note that $left( 0! + 0! + 0! + 0! right)!! = left( 1+1+1+1 right)!! = left (4 right)!! = 8$.
This works and is valid because
The question says I’m allowed to use any of the following symbols in my answer, I am not restricted to using $!$ as an operation only.
edited Sep 7 '18 at 6:42
Laurel
818310
answered Sep 7 '18 at 0:40
El-GuestEl-Guest
21.6k25092
$endgroup$
65
$begingroup$
for those who are confused, !! is a mathematical operator that gives the product of all positive integers upto the argument which have the same parity (odd/even) as the argument, its called a semifactorial. (thus 4!! is 2*4 = 8)
$endgroup$
– casualcoder
Sep 7 '18 at 7:16
13
$begingroup$
@casualcoder most people assume $n!!$ for an arbitrary non-negative integer $n$ is the same as $(n!)!$, but it's not. That brings up some confusion, so I usually write $n!_k$ to denote $n$ with $k$ factorials, but not on this site.
$endgroup$
– user477343
Sep 7 '18 at 11:49
3
$begingroup$
@casualcoder Google disagrees with Wolfram on this.
$endgroup$
– user1717828
Sep 7 '18 at 14:46
10
$begingroup$
Google disagrees with Wolfram on a lot of things. When it comes to math, chances are Wolfram is the correct one.
$endgroup$
– AlexanderJ93
Sep 7 '18 at 23:03
2
$begingroup$
@user1717828, user477343’s comment above explains why Google’s interpretation is not correct in this context.
$endgroup$
– El-Guest
Sep 7 '18 at 23:09
|
show 3 more comments
96
96
96
$begingroup$
I think that
$left( 0! + 0! + 0! + 0! right)!! = 8$.
This is because
$0! = 1$ and $4!! = 8$. Note that $left( 0! + 0! + 0! + 0! right)!! = left( 1+1+1+1 right)!! = left (4 right)!! = 8$.
This works and is valid because
The question says I’m allowed to use any of the following symbols in my answer, I am not restricted to using $!$ as an operation only.
edited Sep 7 '18 at 6:42
Laurel
818310
answered Sep 7 '18 at 0:40
El-GuestEl-Guest
21.6k25092
$endgroup$
I think that
$left( 0! + 0! + 0! + 0! right)!! = 8$.
This is because
$0! = 1$ and $4!! = 8$. Note that $left( 0! + 0! + 0! + 0! right)!! = left( 1+1+1+1 right)!! = left (4 right)!! = 8$.
This works and is valid because
The question says I’m allowed to use any of the following symbols in my answer, I am not restricted to using $!$ as an operation only.
edited Sep 7 '18 at 6:42
Laurel
818310
answered Sep 7 '18 at 0:40
El-GuestEl-Guest
21.6k25092
edited Sep 7 '18 at 6:42
Laurel
818310
edited Sep 7 '18 at 6:42
Laurel
818310
edited Sep 7 '18 at 6:42
Laurel
818310
818310
answered Sep 7 '18 at 0:40
El-GuestEl-Guest
21.6k25092
answered Sep 7 '18 at 0:40
El-GuestEl-Guest
21.6k25092
answered Sep 7 '18 at 0:40
El-GuestEl-Guest
21.6k25092
21.6k25092
65
$begingroup$
for those who are confused, !! is a mathematical operator that gives the product of all positive integers upto the argument which have the same parity (odd/even) as the argument, its called a semifactorial. (thus 4!! is 2*4 = 8)
$endgroup$
– casualcoder
Sep 7 '18 at 7:16
13
$begingroup$
@casualcoder most people assume $n!!$ for an arbitrary non-negative integer $n$ is the same as $(n!)!$, but it's not. That brings up some confusion, so I usually write $n!_k$ to denote $n$ with $k$ factorials, but not on this site.
$endgroup$
– user477343
Sep 7 '18 at 11:49
3
$begingroup$
@casualcoder Google disagrees with Wolfram on this.
$endgroup$
– user1717828
Sep 7 '18 at 14:46
10
$begingroup$
Google disagrees with Wolfram on a lot of things. When it comes to math, chances are Wolfram is the correct one.
$endgroup$
– AlexanderJ93
Sep 7 '18 at 23:03
2
$begingroup$
@user1717828, user477343’s comment above explains why Google’s interpretation is not correct in this context.
$endgroup$
– El-Guest
Sep 7 '18 at 23:09
|
show 3 more comments
65
$begingroup$
for those who are confused, !! is a mathematical operator that gives the product of all positive integers upto the argument which have the same parity (odd/even) as the argument, its called a semifactorial. (thus 4!! is 2*4 = 8)
$endgroup$
– casualcoder
Sep 7 '18 at 7:16
13
$begingroup$
@casualcoder most people assume $n!!$ for an arbitrary non-negative integer $n$ is the same as $(n!)!$, but it's not. That brings up some confusion, so I usually write $n!_k$ to denote $n$ with $k$ factorials, but not on this site.
$endgroup$
– user477343
Sep 7 '18 at 11:49
3
$begingroup$
@casualcoder Google disagrees with Wolfram on this.
$endgroup$
– user1717828
Sep 7 '18 at 14:46
10
$begingroup$
Google disagrees with Wolfram on a lot of things. When it comes to math, chances are Wolfram is the correct one.
$endgroup$
– AlexanderJ93
Sep 7 '18 at 23:03
2
$begingroup$
@user1717828, user477343’s comment above explains why Google’s interpretation is not correct in this context.
$endgroup$
– El-Guest
Sep 7 '18 at 23:09
65
65
$begingroup$
for those who are confused, !! is a mathematical operator that gives the product of all positive integers upto the argument which have the same parity (odd/even) as the argument, its called a semifactorial. (thus 4!! is 2*4 = 8)
$endgroup$
– casualcoder
Sep 7 '18 at 7:16
$begingroup$
for those who are confused, !! is a mathematical operator that gives the product of all positive integers upto the argument which have the same parity (odd/even) as the argument, its called a semifactorial. (thus 4!! is 2*4 = 8)
$endgroup$
– casualcoder
Sep 7 '18 at 7:16
13
13
$begingroup$
@casualcoder most people assume $n!!$ for an arbitrary non-negative integer $n$ is the same as $(n!)!$, but it's not. That brings up some confusion, so I usually write $n!_k$ to denote $n$ with $k$ factorials, but not on this site.
$endgroup$
– user477343
Sep 7 '18 at 11:49
$begingroup$
@casualcoder most people assume $n!!$ for an arbitrary non-negative integer $n$ is the same as $(n!)!$, but it's not. That brings up some confusion, so I usually write $n!_k$ to denote $n$ with $k$ factorials, but not on this site.
$endgroup$
– user477343
Sep 7 '18 at 11:49
3
3
$begingroup$
@casualcoder Google disagrees with Wolfram on this.
$endgroup$
– user1717828
Sep 7 '18 at 14:46
$begingroup$
@casualcoder Google disagrees with Wolfram on this.
$endgroup$
– user1717828
Sep 7 '18 at 14:46
10
10
$begingroup$
Google disagrees with Wolfram on a lot of things. When it comes to math, chances are Wolfram is the correct one.
$endgroup$
– AlexanderJ93
Sep 7 '18 at 23:03
$begingroup$
Google disagrees with Wolfram on a lot of things. When it comes to math, chances are Wolfram is the correct one.
$endgroup$
– AlexanderJ93
Sep 7 '18 at 23:03
2
2
$begingroup$
@user1717828, user477343’s comment above explains why Google’s interpretation is not correct in this context.
$endgroup$
– El-Guest
Sep 7 '18 at 23:09
$begingroup$
@user1717828, user477343’s comment above explains why Google’s interpretation is not correct in this context.
$endgroup$
– El-Guest
Sep 7 '18 at 23:09
|
show 3 more comments
102
$begingroup$
A lateral thinking answer:
0! 0 0 0, because the binary equivalent of 8 is 1000 :)
edited Sep 7 '18 at 14:01
El-Guest
21.6k25092
answered Sep 7 '18 at 13:44
let_the_coding_beginlet_the_coding_begin
1,051114
$endgroup$
9
$begingroup$
I like this! very direct and minimal.
$endgroup$
– Ruadhan2300
Sep 7 '18 at 14:25
2
$begingroup$
My favourite one! I did wonder if someone would go binary.
$endgroup$
– oliver-clare
Sep 7 '18 at 15:07
$begingroup$
I was just checking existing answers to see if anyone else had thought of ! In the sense of negation, and binary. So "yes" :)
$endgroup$
– Stilez
Sep 10 '18 at 8:24
$begingroup$
I have reached my daily voting limit (DVL), but Imma vote this when I can, get it to $99$, and then hope that another user actually puts another upvote :P
$endgroup$
– user477343
Oct 3 '18 at 7:11
add a comment |
102
$begingroup$
A lateral thinking answer:
0! 0 0 0, because the binary equivalent of 8 is 1000 :)
edited Sep 7 '18 at 14:01
El-Guest
21.6k25092
answered Sep 7 '18 at 13:44
let_the_coding_beginlet_the_coding_begin
1,051114
$endgroup$
9
$begingroup$
I like this! very direct and minimal.
$endgroup$
– Ruadhan2300
Sep 7 '18 at 14:25
2
$begingroup$
My favourite one! I did wonder if someone would go binary.
$endgroup$
– oliver-clare
Sep 7 '18 at 15:07
$begingroup$
I was just checking existing answers to see if anyone else had thought of ! In the sense of negation, and binary. So "yes" :)
$endgroup$
– Stilez
Sep 10 '18 at 8:24
$begingroup$
I have reached my daily voting limit (DVL), but Imma vote this when I can, get it to $99$, and then hope that another user actually puts another upvote :P
$endgroup$
– user477343
Oct 3 '18 at 7:11
add a comment |
102
102
102
$begingroup$
A lateral thinking answer:
0! 0 0 0, because the binary equivalent of 8 is 1000 :)
edited Sep 7 '18 at 14:01
El-Guest
21.6k25092
answered Sep 7 '18 at 13:44
let_the_coding_beginlet_the_coding_begin
1,051114
$endgroup$
A lateral thinking answer:
0! 0 0 0, because the binary equivalent of 8 is 1000 :)
edited Sep 7 '18 at 14:01
El-Guest
21.6k25092
answered Sep 7 '18 at 13:44
let_the_coding_beginlet_the_coding_begin
1,051114
edited Sep 7 '18 at 14:01
El-Guest
21.6k25092
edited Sep 7 '18 at 14:01
El-Guest
21.6k25092
edited Sep 7 '18 at 14:01
El-Guest
21.6k25092
21.6k25092
answered Sep 7 '18 at 13:44
let_the_coding_beginlet_the_coding_begin
1,051114
answered Sep 7 '18 at 13:44
let_the_coding_beginlet_the_coding_begin
1,051114
answered Sep 7 '18 at 13:44
let_the_coding_beginlet_the_coding_begin
1,051114
1,051114
9
$begingroup$
I like this! very direct and minimal.
$endgroup$
– Ruadhan2300
Sep 7 '18 at 14:25
2
$begingroup$
My favourite one! I did wonder if someone would go binary.
$endgroup$
– oliver-clare
Sep 7 '18 at 15:07
$begingroup$
I was just checking existing answers to see if anyone else had thought of ! In the sense of negation, and binary. So "yes" :)
$endgroup$
– Stilez
Sep 10 '18 at 8:24
$begingroup$
I have reached my daily voting limit (DVL), but Imma vote this when I can, get it to $99$, and then hope that another user actually puts another upvote :P
$endgroup$
– user477343
Oct 3 '18 at 7:11
add a comment |
9
$begingroup$
I like this! very direct and minimal.
$endgroup$
– Ruadhan2300
Sep 7 '18 at 14:25
2
$begingroup$
My favourite one! I did wonder if someone would go binary.
$endgroup$
– oliver-clare
Sep 7 '18 at 15:07
$begingroup$
I was just checking existing answers to see if anyone else had thought of ! In the sense of negation, and binary. So "yes" :)
$endgroup$
– Stilez
Sep 10 '18 at 8:24
$begingroup$
I have reached my daily voting limit (DVL), but Imma vote this when I can, get it to $99$, and then hope that another user actually puts another upvote :P
$endgroup$
– user477343
Oct 3 '18 at 7:11
9
9
$begingroup$
I like this! very direct and minimal.
$endgroup$
– Ruadhan2300
Sep 7 '18 at 14:25
$begingroup$
I like this! very direct and minimal.
$endgroup$
– Ruadhan2300
Sep 7 '18 at 14:25
2
2
$begingroup$
My favourite one! I did wonder if someone would go binary.
$endgroup$
– oliver-clare
Sep 7 '18 at 15:07
$begingroup$
My favourite one! I did wonder if someone would go binary.
$endgroup$
– oliver-clare
Sep 7 '18 at 15:07
$begingroup$
I was just checking existing answers to see if anyone else had thought of ! In the sense of negation, and binary. So "yes" :)
$endgroup$
– Stilez
Sep 10 '18 at 8:24
$begingroup$
I was just checking existing answers to see if anyone else had thought of ! In the sense of negation, and binary. So "yes" :)
$endgroup$
– Stilez
Sep 10 '18 at 8:24
$begingroup$
I have reached my daily voting limit (DVL), but Imma vote this when I can, get it to $99$, and then hope that another user actually puts another upvote :P
$endgroup$
– user477343
Oct 3 '18 at 7:11
$begingroup$
I have reached my daily voting limit (DVL), but Imma vote this when I can, get it to $99$, and then hope that another user actually puts another upvote :P
$endgroup$
– user477343
Oct 3 '18 at 7:11
add a comment |
68
$begingroup$
$0 + 0 + 0 + 0 ~~!!=~ 8$
because
$ !!= $ is an alternative way of writing $ ne $.
answered Sep 7 '18 at 8:43
Teemu PiippoTeemu Piippo
66724
$endgroup$
4
$begingroup$
This is the answer!
$endgroup$
– user51438
Sep 8 '18 at 2:20
3
$begingroup$
@TheodoreWeld nope, this question is tagged mathematics, and you have to use a slash similar to Yout Ried instead of this
$endgroup$
– phuclv
Sep 8 '18 at 4:24
add a comment |
68
$begingroup$
$0 + 0 + 0 + 0 ~~!!=~ 8$
because
$ !!= $ is an alternative way of writing $ ne $.
answered Sep 7 '18 at 8:43
Teemu PiippoTeemu Piippo
66724
$endgroup$
4
$begingroup$
This is the answer!
$endgroup$
– user51438
Sep 8 '18 at 2:20
3
$begingroup$
@TheodoreWeld nope, this question is tagged mathematics, and you have to use a slash similar to Yout Ried instead of this
$endgroup$
– phuclv
Sep 8 '18 at 4:24
add a comment |
68
68
68
$begingroup$
$0 + 0 + 0 + 0 ~~!!=~ 8$
because
$ !!= $ is an alternative way of writing $ ne $.
answered Sep 7 '18 at 8:43
Teemu PiippoTeemu Piippo
66724
$endgroup$
$0 + 0 + 0 + 0 ~~!!=~ 8$
because
$ !!= $ is an alternative way of writing $ ne $.
answered Sep 7 '18 at 8:43
Teemu PiippoTeemu Piippo
66724
answered Sep 7 '18 at 8:43
Teemu PiippoTeemu Piippo
66724
answered Sep 7 '18 at 8:43
Teemu PiippoTeemu Piippo
66724
answered Sep 7 '18 at 8:43
Teemu PiippoTeemu Piippo
66724
66724
4
$begingroup$
This is the answer!
$endgroup$
– user51438
Sep 8 '18 at 2:20
3
$begingroup$
@TheodoreWeld nope, this question is tagged mathematics, and you have to use a slash similar to Yout Ried instead of this
$endgroup$
– phuclv
Sep 8 '18 at 4:24
add a comment |
4
$begingroup$
This is the answer!
$endgroup$
– user51438
Sep 8 '18 at 2:20
3
$begingroup$
@TheodoreWeld nope, this question is tagged mathematics, and you have to use a slash similar to Yout Ried instead of this
$endgroup$
– phuclv
Sep 8 '18 at 4:24
4
4
$begingroup$
This is the answer!
$endgroup$
– user51438
Sep 8 '18 at 2:20
$begingroup$
This is the answer!
$endgroup$
– user51438
Sep 8 '18 at 2:20
3
3
$begingroup$
@TheodoreWeld nope, this question is tagged mathematics, and you have to use a slash similar to Yout Ried instead of this
$endgroup$
– phuclv
Sep 8 '18 at 4:24
$begingroup$
@TheodoreWeld nope, this question is tagged mathematics, and you have to use a slash similar to Yout Ried instead of this
$endgroup$
– phuclv
Sep 8 '18 at 4:24
add a comment |
55
$begingroup$
Lateral thinking!
$$0+0+substack{0\0}=0+0+8=8$$
answered Sep 7 '18 at 6:25
TheSimpliFireTheSimpliFire
2,155532
$endgroup$
8
$begingroup$
First I thought "how does picking 0 out of 0 give you 8 combinations", then I got it.
$endgroup$
– Fabian Röling
Sep 7 '18 at 12:01
82
$begingroup$
Looks more like vertical thinking to me.
$endgroup$
– Evargalo
Sep 7 '18 at 12:02
add a comment |
55
$begingroup$
Lateral thinking!
$$0+0+substack{0\0}=0+0+8=8$$
answered Sep 7 '18 at 6:25
TheSimpliFireTheSimpliFire
2,155532
$endgroup$
8
$begingroup$
First I thought "how does picking 0 out of 0 give you 8 combinations", then I got it.
$endgroup$
– Fabian Röling
Sep 7 '18 at 12:01
82
$begingroup$
Looks more like vertical thinking to me.
$endgroup$
– Evargalo
Sep 7 '18 at 12:02
add a comment |
55
55
55
$begingroup$
Lateral thinking!
$$0+0+substack{0\0}=0+0+8=8$$
answered Sep 7 '18 at 6:25
TheSimpliFireTheSimpliFire
2,155532
$endgroup$
Lateral thinking!
$$0+0+substack{0\0}=0+0+8=8$$
answered Sep 7 '18 at 6:25
TheSimpliFireTheSimpliFire
2,155532
answered Sep 7 '18 at 6:25
TheSimpliFireTheSimpliFire
2,155532
answered Sep 7 '18 at 6:25
TheSimpliFireTheSimpliFire
2,155532
answered Sep 7 '18 at 6:25
TheSimpliFireTheSimpliFire
2,155532
2,155532
8
$begingroup$
First I thought "how does picking 0 out of 0 give you 8 combinations", then I got it.
$endgroup$
– Fabian Röling
Sep 7 '18 at 12:01
82
$begingroup$
Looks more like vertical thinking to me.
$endgroup$
– Evargalo
Sep 7 '18 at 12:02
add a comment |
8
$begingroup$
First I thought "how does picking 0 out of 0 give you 8 combinations", then I got it.
$endgroup$
– Fabian Röling
Sep 7 '18 at 12:01
82
$begingroup$
Looks more like vertical thinking to me.
$endgroup$
– Evargalo
Sep 7 '18 at 12:02
8
8
$begingroup$
First I thought "how does picking 0 out of 0 give you 8 combinations", then I got it.
$endgroup$
– Fabian Röling
Sep 7 '18 at 12:01
$begingroup$
First I thought "how does picking 0 out of 0 give you 8 combinations", then I got it.
$endgroup$
– Fabian Röling
Sep 7 '18 at 12:01
82
82
$begingroup$
Looks more like vertical thinking to me.
$endgroup$
– Evargalo
Sep 7 '18 at 12:02
$begingroup$
Looks more like vertical thinking to me.
$endgroup$
– Evargalo
Sep 7 '18 at 12:02
add a comment |
34
$begingroup$
let me try:
$0! Vert 0 - 0!-0! =8$
$10-1-1=8$
$Vert$ is a concatenation operation
answered Sep 7 '18 at 0:43
malioboromalioboro
2,77311033
$endgroup$
1
$begingroup$
good solution but in this case you cant use '[' or ']'
$endgroup$
– casualcoder
Sep 7 '18 at 7:13
3
$begingroup$
@casualcoder It was explicitly allowed. If this wasn't meant to be a valid solution, why even allow it, as 00 = 0+0 = 0?
$endgroup$
– Zizy Archer
Sep 7 '18 at 7:51
1
$begingroup$
I've never used $[ldots ]$ as an operator before, let alone for what you use it as in this answer. I usually use it as a set notation; i.e., $[n]={1,2,ldots n}$ :P
$endgroup$
– user477343
Sep 7 '18 at 11:53
2
$begingroup$
I think "concatenation allowed" in these puzzles usually means that, for example,2 8 7can be28 ÷ 7 = 4, but not(2+8)7 = 107
$endgroup$
– Chronocidal
Sep 7 '18 at 14:22
3
$begingroup$
@Chronocidal What you are saying is probably true about most of these puzzles. But in this particular puzzle where all the digits are 0, that interpretation does not make a lot of sense. To me this answer feels more legitimate than the accepted answer. That's because concatenation is explicitly allowed, but the !! is not explicitly allowed (though it may be implied).
$endgroup$
– kasperd
Sep 8 '18 at 12:20
|
show 3 more comments
34
$begingroup$
let me try:
$0! Vert 0 - 0!-0! =8$
$10-1-1=8$
$Vert$ is a concatenation operation
answered Sep 7 '18 at 0:43
malioboromalioboro
2,77311033
$endgroup$
1
$begingroup$
good solution but in this case you cant use '[' or ']'
$endgroup$
– casualcoder
Sep 7 '18 at 7:13
3
$begingroup$
@casualcoder It was explicitly allowed. If this wasn't meant to be a valid solution, why even allow it, as 00 = 0+0 = 0?
$endgroup$
– Zizy Archer
Sep 7 '18 at 7:51
1
$begingroup$
I've never used $[ldots ]$ as an operator before, let alone for what you use it as in this answer. I usually use it as a set notation; i.e., $[n]={1,2,ldots n}$ :P
$endgroup$
– user477343
Sep 7 '18 at 11:53
2
$begingroup$
I think "concatenation allowed" in these puzzles usually means that, for example,2 8 7can be28 ÷ 7 = 4, but not(2+8)7 = 107
$endgroup$
– Chronocidal
Sep 7 '18 at 14:22
3
$begingroup$
@Chronocidal What you are saying is probably true about most of these puzzles. But in this particular puzzle where all the digits are 0, that interpretation does not make a lot of sense. To me this answer feels more legitimate than the accepted answer. That's because concatenation is explicitly allowed, but the !! is not explicitly allowed (though it may be implied).
$endgroup$
– kasperd
Sep 8 '18 at 12:20
|
show 3 more comments
34
34
34
$begingroup$
let me try:
$0! Vert 0 - 0!-0! =8$
$10-1-1=8$
$Vert$ is a concatenation operation
answered Sep 7 '18 at 0:43
malioboromalioboro
2,77311033
$endgroup$
let me try:
$0! Vert 0 - 0!-0! =8$
$10-1-1=8$
$Vert$ is a concatenation operation
answered Sep 7 '18 at 0:43
malioboromalioboro
2,77311033
edited Sep 8 '18 at 15:29
answered Sep 7 '18 at 0:43
malioboromalioboro
2,77311033
answered Sep 7 '18 at 0:43
malioboromalioboro
2,77311033
answered Sep 7 '18 at 0:43
malioboromalioboro
2,77311033
2,77311033
1
$begingroup$
good solution but in this case you cant use '[' or ']'
$endgroup$
– casualcoder
Sep 7 '18 at 7:13
3
$begingroup$
@casualcoder It was explicitly allowed. If this wasn't meant to be a valid solution, why even allow it, as 00 = 0+0 = 0?
$endgroup$
– Zizy Archer
Sep 7 '18 at 7:51
1
$begingroup$
I've never used $[ldots ]$ as an operator before, let alone for what you use it as in this answer. I usually use it as a set notation; i.e., $[n]={1,2,ldots n}$ :P
$endgroup$
– user477343
Sep 7 '18 at 11:53
2
$begingroup$
I think "concatenation allowed" in these puzzles usually means that, for example,2 8 7can be28 ÷ 7 = 4, but not(2+8)7 = 107
$endgroup$
– Chronocidal
Sep 7 '18 at 14:22
3
$begingroup$
@Chronocidal What you are saying is probably true about most of these puzzles. But in this particular puzzle where all the digits are 0, that interpretation does not make a lot of sense. To me this answer feels more legitimate than the accepted answer. That's because concatenation is explicitly allowed, but the !! is not explicitly allowed (though it may be implied).
$endgroup$
– kasperd
Sep 8 '18 at 12:20
|
show 3 more comments
1
$begingroup$
good solution but in this case you cant use '[' or ']'
$endgroup$
– casualcoder
Sep 7 '18 at 7:13
3
$begingroup$
@casualcoder It was explicitly allowed. If this wasn't meant to be a valid solution, why even allow it, as 00 = 0+0 = 0?
$endgroup$
– Zizy Archer
Sep 7 '18 at 7:51
1
$begingroup$
I've never used $[ldots ]$ as an operator before, let alone for what you use it as in this answer. I usually use it as a set notation; i.e., $[n]={1,2,ldots n}$ :P
$endgroup$
– user477343
Sep 7 '18 at 11:53
2
$begingroup$
I think "concatenation allowed" in these puzzles usually means that, for example,2 8 7can be28 ÷ 7 = 4, but not(2+8)7 = 107
$endgroup$
– Chronocidal
Sep 7 '18 at 14:22
3
$begingroup$
@Chronocidal What you are saying is probably true about most of these puzzles. But in this particular puzzle where all the digits are 0, that interpretation does not make a lot of sense. To me this answer feels more legitimate than the accepted answer. That's because concatenation is explicitly allowed, but the !! is not explicitly allowed (though it may be implied).
$endgroup$
– kasperd
Sep 8 '18 at 12:20
1
1
$begingroup$
good solution but in this case you cant use '[' or ']'
$endgroup$
– casualcoder
Sep 7 '18 at 7:13
$begingroup$
good solution but in this case you cant use '[' or ']'
$endgroup$
– casualcoder
Sep 7 '18 at 7:13
3
3
$begingroup$
@casualcoder It was explicitly allowed. If this wasn't meant to be a valid solution, why even allow it, as 00 = 0+0 = 0?
$endgroup$
– Zizy Archer
Sep 7 '18 at 7:51
$begingroup$
@casualcoder It was explicitly allowed. If this wasn't meant to be a valid solution, why even allow it, as 00 = 0+0 = 0?
$endgroup$
– Zizy Archer
Sep 7 '18 at 7:51
1
1
$begingroup$
I've never used $[ldots ]$ as an operator before, let alone for what you use it as in this answer. I usually use it as a set notation; i.e., $[n]={1,2,ldots n}$ :P
$endgroup$
– user477343
Sep 7 '18 at 11:53
$begingroup$
I've never used $[ldots ]$ as an operator before, let alone for what you use it as in this answer. I usually use it as a set notation; i.e., $[n]={1,2,ldots n}$ :P
$endgroup$
– user477343
Sep 7 '18 at 11:53
2
2
$begingroup$
I think "concatenation allowed" in these puzzles usually means that, for example,
2 8 7 can be 28 ÷ 7 = 4, but not (2+8)7 = 107$endgroup$
– Chronocidal
Sep 7 '18 at 14:22
$begingroup$
I think "concatenation allowed" in these puzzles usually means that, for example,
2 8 7 can be 28 ÷ 7 = 4, but not (2+8)7 = 107$endgroup$
– Chronocidal
Sep 7 '18 at 14:22
3
3
$begingroup$
@Chronocidal What you are saying is probably true about most of these puzzles. But in this particular puzzle where all the digits are 0, that interpretation does not make a lot of sense. To me this answer feels more legitimate than the accepted answer. That's because concatenation is explicitly allowed, but the !! is not explicitly allowed (though it may be implied).
$endgroup$
– kasperd
Sep 8 '18 at 12:20
$begingroup$
@Chronocidal What you are saying is probably true about most of these puzzles. But in this particular puzzle where all the digits are 0, that interpretation does not make a lot of sense. To me this answer feels more legitimate than the accepted answer. That's because concatenation is explicitly allowed, but the !! is not explicitly allowed (though it may be implied).
$endgroup$
– kasperd
Sep 8 '18 at 12:20
|
show 3 more comments
29
$begingroup$
$((0!+0!)^{(0!+0!)})!!$
Evaluation:
$((0!+0!)^{(0!+0!)})!!$
$rightarrow ((1+1)^{(1+1)})!!$
$rightarrow (2^2)!!$
$rightarrow 4!! = 8$
answered Sep 7 '18 at 14:04
u_ndefinedu_ndefined
2,7801437
$endgroup$
$begingroup$
How did I not think of that?? DVL16 :
$endgroup$
– user477343
Oct 3 '18 at 7:12
add a comment |
29
$begingroup$
$((0!+0!)^{(0!+0!)})!!$
Evaluation:
$((0!+0!)^{(0!+0!)})!!$
$rightarrow ((1+1)^{(1+1)})!!$
$rightarrow (2^2)!!$
$rightarrow 4!! = 8$
answered Sep 7 '18 at 14:04
u_ndefinedu_ndefined
2,7801437
$endgroup$
$begingroup$
How did I not think of that?? DVL16 :
$endgroup$
– user477343
Oct 3 '18 at 7:12
add a comment |
29
29
29
$begingroup$
$((0!+0!)^{(0!+0!)})!!$
Evaluation:
$((0!+0!)^{(0!+0!)})!!$
$rightarrow ((1+1)^{(1+1)})!!$
$rightarrow (2^2)!!$
$rightarrow 4!! = 8$
answered Sep 7 '18 at 14:04
u_ndefinedu_ndefined
2,7801437
$endgroup$
$((0!+0!)^{(0!+0!)})!!$
Evaluation:
$((0!+0!)^{(0!+0!)})!!$
$rightarrow ((1+1)^{(1+1)})!!$
$rightarrow (2^2)!!$
$rightarrow 4!! = 8$
answered Sep 7 '18 at 14:04
u_ndefinedu_ndefined
2,7801437
answered Sep 7 '18 at 14:04
u_ndefinedu_ndefined
2,7801437
answered Sep 7 '18 at 14:04
u_ndefinedu_ndefined
2,7801437
answered Sep 7 '18 at 14:04
u_ndefinedu_ndefined
2,7801437
2,7801437
$begingroup$
How did I not think of that?? DVL16 :
$endgroup$
– user477343
Oct 3 '18 at 7:12
add a comment |
$begingroup$
How did I not think of that?? DVL16 :
$endgroup$
– user477343
Oct 3 '18 at 7:12
$begingroup$
How did I not think of that?? DVL16 :
$endgroup$
– user477343
Oct 3 '18 at 7:12
$begingroup$
How did I not think of that?? DVL16 :
$endgroup$
– user477343
Oct 3 '18 at 7:12
add a comment |
27
$begingroup$
It's different:
$,++$
$0;;;0$
$,++$
$0;;;0$
$,++$
An ASCII art $8$ using only four $0$'s and $+$'s.
answered Sep 7 '18 at 14:46
JonMark PerryJonMark Perry
20.6k64099
$endgroup$
add a comment |
27
$begingroup$
It's different:
$,++$
$0;;;0$
$,++$
$0;;;0$
$,++$
An ASCII art $8$ using only four $0$'s and $+$'s.
answered Sep 7 '18 at 14:46
JonMark PerryJonMark Perry
20.6k64099
$endgroup$
add a comment |
27
27
27
$begingroup$
It's different:
$,++$
$0;;;0$
$,++$
$0;;;0$
$,++$
An ASCII art $8$ using only four $0$'s and $+$'s.
answered Sep 7 '18 at 14:46
JonMark PerryJonMark Perry
20.6k64099
$endgroup$
It's different:
$,++$
$0;;;0$
$,++$
$0;;;0$
$,++$
An ASCII art $8$ using only four $0$'s and $+$'s.
answered Sep 7 '18 at 14:46
JonMark PerryJonMark Perry
20.6k64099
answered Sep 7 '18 at 14:46
JonMark PerryJonMark Perry
20.6k64099
answered Sep 7 '18 at 14:46
JonMark PerryJonMark Perry
20.6k64099
answered Sep 7 '18 at 14:46
JonMark PerryJonMark Perry
20.6k64099
20.6k64099
add a comment |
add a comment |
16
$begingroup$
0 + 0 + 0 + 0 = !8
because
In C/C++, ! refers to the logical not operator, where all non-zero values become 0, and 0 becomes 1.
answered Sep 7 '18 at 22:48
l kl k
1612
$endgroup$
$begingroup$
I think this should be "the binary not operator".
$endgroup$
– Raimund Krämer
Sep 11 '18 at 8:17
3
$begingroup$
@RaimundKrämer ITYM unary not. C99 calls it the "logical negation operator" and files it in the chapter "Unary operators".
$endgroup$
– Jens
Sep 11 '18 at 15:56
$begingroup$
But !8 can be a subfactorial of 8.
$endgroup$
– rus9384
Sep 12 '18 at 11:39
$begingroup$
@Raimund Krämer, In programming, we have logical operators and bitwise operators. Logical operators (!,&&,||in C) operate on boolean values (true and false). Bitwise operators (~,&,|,^in C) operate on bits (0 and 1) of numbers. Calling either set binary operators would be confusing, as both set work on binary values.
$endgroup$
– ikegami
Sep 13 '18 at 11:12
$begingroup$
@Raimund Krämer Also, "binary operator" can mean any operator on 2 values (as opposed to a unary operator such as ! or ~)
$endgroup$
– l k
Sep 18 '18 at 6:23
|
show 1 more comment
16
$begingroup$
0 + 0 + 0 + 0 = !8
because
In C/C++, ! refers to the logical not operator, where all non-zero values become 0, and 0 becomes 1.
answered Sep 7 '18 at 22:48
l kl k
1612
$endgroup$
$begingroup$
I think this should be "the binary not operator".
$endgroup$
– Raimund Krämer
Sep 11 '18 at 8:17
3
$begingroup$
@RaimundKrämer ITYM unary not. C99 calls it the "logical negation operator" and files it in the chapter "Unary operators".
$endgroup$
– Jens
Sep 11 '18 at 15:56
$begingroup$
But !8 can be a subfactorial of 8.
$endgroup$
– rus9384
Sep 12 '18 at 11:39
$begingroup$
@Raimund Krämer, In programming, we have logical operators and bitwise operators. Logical operators (!,&&,||in C) operate on boolean values (true and false). Bitwise operators (~,&,|,^in C) operate on bits (0 and 1) of numbers. Calling either set binary operators would be confusing, as both set work on binary values.
$endgroup$
– ikegami
Sep 13 '18 at 11:12
$begingroup$
@Raimund Krämer Also, "binary operator" can mean any operator on 2 values (as opposed to a unary operator such as ! or ~)
$endgroup$
– l k
Sep 18 '18 at 6:23
|
show 1 more comment
16
16
16
$begingroup$
0 + 0 + 0 + 0 = !8
because
In C/C++, ! refers to the logical not operator, where all non-zero values become 0, and 0 becomes 1.
answered Sep 7 '18 at 22:48
l kl k
1612
$endgroup$
0 + 0 + 0 + 0 = !8
because
In C/C++, ! refers to the logical not operator, where all non-zero values become 0, and 0 becomes 1.
answered Sep 7 '18 at 22:48
l kl k
1612
answered Sep 7 '18 at 22:48
l kl k
1612
answered Sep 7 '18 at 22:48
l kl k
1612
answered Sep 7 '18 at 22:48
l kl k
1612
1612
$begingroup$
I think this should be "the binary not operator".
$endgroup$
– Raimund Krämer
Sep 11 '18 at 8:17
3
$begingroup$
@RaimundKrämer ITYM unary not. C99 calls it the "logical negation operator" and files it in the chapter "Unary operators".
$endgroup$
– Jens
Sep 11 '18 at 15:56
$begingroup$
But !8 can be a subfactorial of 8.
$endgroup$
– rus9384
Sep 12 '18 at 11:39
$begingroup$
@Raimund Krämer, In programming, we have logical operators and bitwise operators. Logical operators (!,&&,||in C) operate on boolean values (true and false). Bitwise operators (~,&,|,^in C) operate on bits (0 and 1) of numbers. Calling either set binary operators would be confusing, as both set work on binary values.
$endgroup$
– ikegami
Sep 13 '18 at 11:12
$begingroup$
@Raimund Krämer Also, "binary operator" can mean any operator on 2 values (as opposed to a unary operator such as ! or ~)
$endgroup$
– l k
Sep 18 '18 at 6:23
|
show 1 more comment
$begingroup$
I think this should be "the binary not operator".
$endgroup$
– Raimund Krämer
Sep 11 '18 at 8:17
3
$begingroup$
@RaimundKrämer ITYM unary not. C99 calls it the "logical negation operator" and files it in the chapter "Unary operators".
$endgroup$
– Jens
Sep 11 '18 at 15:56
$begingroup$
But !8 can be a subfactorial of 8.
$endgroup$
– rus9384
Sep 12 '18 at 11:39
$begingroup$
@Raimund Krämer, In programming, we have logical operators and bitwise operators. Logical operators (!,&&,||in C) operate on boolean values (true and false). Bitwise operators (~,&,|,^in C) operate on bits (0 and 1) of numbers. Calling either set binary operators would be confusing, as both set work on binary values.
$endgroup$
– ikegami
Sep 13 '18 at 11:12
$begingroup$
@Raimund Krämer Also, "binary operator" can mean any operator on 2 values (as opposed to a unary operator such as ! or ~)
$endgroup$
– l k
Sep 18 '18 at 6:23
$begingroup$
I think this should be "the binary not operator".
$endgroup$
– Raimund Krämer
Sep 11 '18 at 8:17
$begingroup$
I think this should be "the binary not operator".
$endgroup$
– Raimund Krämer
Sep 11 '18 at 8:17
3
3
$begingroup$
@RaimundKrämer ITYM unary not. C99 calls it the "logical negation operator" and files it in the chapter "Unary operators".
$endgroup$
– Jens
Sep 11 '18 at 15:56
$begingroup$
@RaimundKrämer ITYM unary not. C99 calls it the "logical negation operator" and files it in the chapter "Unary operators".
$endgroup$
– Jens
Sep 11 '18 at 15:56
$begingroup$
But !8 can be a subfactorial of 8.
$endgroup$
– rus9384
Sep 12 '18 at 11:39
$begingroup$
But !8 can be a subfactorial of 8.
$endgroup$
– rus9384
Sep 12 '18 at 11:39
$begingroup$
@Raimund Krämer, In programming, we have logical operators and bitwise operators. Logical operators (
!, &&, || in C) operate on boolean values (true and false). Bitwise operators (~, &, |, ^ in C) operate on bits (0 and 1) of numbers. Calling either set binary operators would be confusing, as both set work on binary values.$endgroup$
– ikegami
Sep 13 '18 at 11:12
$begingroup$
@Raimund Krämer, In programming, we have logical operators and bitwise operators. Logical operators (
!, &&, || in C) operate on boolean values (true and false). Bitwise operators (~, &, |, ^ in C) operate on bits (0 and 1) of numbers. Calling either set binary operators would be confusing, as both set work on binary values.$endgroup$
– ikegami
Sep 13 '18 at 11:12
$begingroup$
@Raimund Krämer Also, "binary operator" can mean any operator on 2 values (as opposed to a unary operator such as ! or ~)
$endgroup$
– l k
Sep 18 '18 at 6:23
$begingroup$
@Raimund Krämer Also, "binary operator" can mean any operator on 2 values (as opposed to a unary operator such as ! or ~)
$endgroup$
– l k
Sep 18 '18 at 6:23
|
show 1 more comment
14
$begingroup$
It's just a matter of perspective ...
0!/0 + 0!/0 = ∞
My reasoning....
0/0 is undefined so we have to first change the 0's into 1's with 0!
(...and why did you write the infinity symbol sideways in your question?)
answered Sep 7 '18 at 18:20
rrauenzarrauenza
24915
$endgroup$
1
$begingroup$
x/0 isn't infinite, though...
$endgroup$
– Adam Smith
Sep 8 '18 at 17:49
$begingroup$
Lateral thinking was yesterday. Vertical thinking is the new kid in town!
$endgroup$
– Jens
Sep 11 '18 at 15:59
1
$begingroup$
x/0 == +infinity, per ieee 754
$endgroup$
– j__m
Sep 15 '18 at 12:40
add a comment |
14
$begingroup$
It's just a matter of perspective ...
0!/0 + 0!/0 = ∞
My reasoning....
0/0 is undefined so we have to first change the 0's into 1's with 0!
(...and why did you write the infinity symbol sideways in your question?)
answered Sep 7 '18 at 18:20
rrauenzarrauenza
24915
$endgroup$
1
$begingroup$
x/0 isn't infinite, though...
$endgroup$
– Adam Smith
Sep 8 '18 at 17:49
$begingroup$
Lateral thinking was yesterday. Vertical thinking is the new kid in town!
$endgroup$
– Jens
Sep 11 '18 at 15:59
1
$begingroup$
x/0 == +infinity, per ieee 754
$endgroup$
– j__m
Sep 15 '18 at 12:40
add a comment |
14
14
14
$begingroup$
It's just a matter of perspective ...
0!/0 + 0!/0 = ∞
My reasoning....
0/0 is undefined so we have to first change the 0's into 1's with 0!
(...and why did you write the infinity symbol sideways in your question?)
answered Sep 7 '18 at 18:20
rrauenzarrauenza
24915
$endgroup$
It's just a matter of perspective ...
0!/0 + 0!/0 = ∞
My reasoning....
0/0 is undefined so we have to first change the 0's into 1's with 0!
(...and why did you write the infinity symbol sideways in your question?)
answered Sep 7 '18 at 18:20
rrauenzarrauenza
24915
edited Sep 8 '18 at 19:31
answered Sep 7 '18 at 18:20
rrauenzarrauenza
24915
answered Sep 7 '18 at 18:20
rrauenzarrauenza
24915
answered Sep 7 '18 at 18:20
rrauenzarrauenza
24915
24915
1
$begingroup$
x/0 isn't infinite, though...
$endgroup$
– Adam Smith
Sep 8 '18 at 17:49
$begingroup$
Lateral thinking was yesterday. Vertical thinking is the new kid in town!
$endgroup$
– Jens
Sep 11 '18 at 15:59
1
$begingroup$
x/0 == +infinity, per ieee 754
$endgroup$
– j__m
Sep 15 '18 at 12:40
add a comment |
1
$begingroup$
x/0 isn't infinite, though...
$endgroup$
– Adam Smith
Sep 8 '18 at 17:49
$begingroup$
Lateral thinking was yesterday. Vertical thinking is the new kid in town!
$endgroup$
– Jens
Sep 11 '18 at 15:59
1
$begingroup$
x/0 == +infinity, per ieee 754
$endgroup$
– j__m
Sep 15 '18 at 12:40
1
1
$begingroup$
x/0 isn't infinite, though...
$endgroup$
– Adam Smith
Sep 8 '18 at 17:49
$begingroup$
x/0 isn't infinite, though...
$endgroup$
– Adam Smith
Sep 8 '18 at 17:49
$begingroup$
Lateral thinking was yesterday. Vertical thinking is the new kid in town!
$endgroup$
– Jens
Sep 11 '18 at 15:59
$begingroup$
Lateral thinking was yesterday. Vertical thinking is the new kid in town!
$endgroup$
– Jens
Sep 11 '18 at 15:59
1
1
$begingroup$
x/0 == +infinity, per ieee 754
$endgroup$
– j__m
Sep 15 '18 at 12:40
$begingroup$
x/0 == +infinity, per ieee 754
$endgroup$
– j__m
Sep 15 '18 at 12:40
add a comment |
10
$begingroup$
$$[+!0]+[0]-!0-!0$$
Works in JavaScript. Hit F12 and type this into the console (This equation editor uses "−" instead of "-" so copy and paste doesn't quite work). Otherwise, it works in the same way as @malioboro and @Arnaldur's answers.
In fact, you can make any JavaScript application run just by using a combination of 6 characters, which is what inspired me to make this. I substituted + for 0 when asking JSF**k to do 10-2.
answered Sep 7 '18 at 12:35
NamytsNamyts
4647
$endgroup$
3
$begingroup$
I like this abuse of javascript a lot. !0 for true, +!0 for 1, [1] for array containing 1, [1]+[0] for concatenate two arrays to get the string "10" ; subtract 1 from that twice to end up as 8. Love it.
$endgroup$
– Ross Presser
Sep 8 '18 at 4:10
2
$begingroup$
but the question is tagged mathematics and not programming
$endgroup$
– phuclv
Sep 8 '18 at 4:21
3
$begingroup$
Who says you can use square brackets....
$endgroup$
– user52269
Sep 8 '18 at 6:57
add a comment |
10
$begingroup$
$$[+!0]+[0]-!0-!0$$
Works in JavaScript. Hit F12 and type this into the console (This equation editor uses "−" instead of "-" so copy and paste doesn't quite work). Otherwise, it works in the same way as @malioboro and @Arnaldur's answers.
In fact, you can make any JavaScript application run just by using a combination of 6 characters, which is what inspired me to make this. I substituted + for 0 when asking JSF**k to do 10-2.
answered Sep 7 '18 at 12:35
NamytsNamyts
4647
$endgroup$
3
$begingroup$
I like this abuse of javascript a lot. !0 for true, +!0 for 1, [1] for array containing 1, [1]+[0] for concatenate two arrays to get the string "10" ; subtract 1 from that twice to end up as 8. Love it.
$endgroup$
– Ross Presser
Sep 8 '18 at 4:10
2
$begingroup$
but the question is tagged mathematics and not programming
$endgroup$
– phuclv
Sep 8 '18 at 4:21
3
$begingroup$
Who says you can use square brackets....
$endgroup$
– user52269
Sep 8 '18 at 6:57
add a comment |
10
10
10
$begingroup$
$$[+!0]+[0]-!0-!0$$
Works in JavaScript. Hit F12 and type this into the console (This equation editor uses "−" instead of "-" so copy and paste doesn't quite work). Otherwise, it works in the same way as @malioboro and @Arnaldur's answers.
In fact, you can make any JavaScript application run just by using a combination of 6 characters, which is what inspired me to make this. I substituted + for 0 when asking JSF**k to do 10-2.
answered Sep 7 '18 at 12:35
NamytsNamyts
4647
$endgroup$
$$[+!0]+[0]-!0-!0$$
Works in JavaScript. Hit F12 and type this into the console (This equation editor uses "−" instead of "-" so copy and paste doesn't quite work). Otherwise, it works in the same way as @malioboro and @Arnaldur's answers.
In fact, you can make any JavaScript application run just by using a combination of 6 characters, which is what inspired me to make this. I substituted + for 0 when asking JSF**k to do 10-2.
answered Sep 7 '18 at 12:35
NamytsNamyts
4647
answered Sep 7 '18 at 12:35
NamytsNamyts
4647
answered Sep 7 '18 at 12:35
NamytsNamyts
4647
answered Sep 7 '18 at 12:35
NamytsNamyts
4647
4647
3
$begingroup$
I like this abuse of javascript a lot. !0 for true, +!0 for 1, [1] for array containing 1, [1]+[0] for concatenate two arrays to get the string "10" ; subtract 1 from that twice to end up as 8. Love it.
$endgroup$
– Ross Presser
Sep 8 '18 at 4:10
2
$begingroup$
but the question is tagged mathematics and not programming
$endgroup$
– phuclv
Sep 8 '18 at 4:21
3
$begingroup$
Who says you can use square brackets....
$endgroup$
– user52269
Sep 8 '18 at 6:57
add a comment |
3
$begingroup$
I like this abuse of javascript a lot. !0 for true, +!0 for 1, [1] for array containing 1, [1]+[0] for concatenate two arrays to get the string "10" ; subtract 1 from that twice to end up as 8. Love it.
$endgroup$
– Ross Presser
Sep 8 '18 at 4:10
2
$begingroup$
but the question is tagged mathematics and not programming
$endgroup$
– phuclv
Sep 8 '18 at 4:21
3
$begingroup$
Who says you can use square brackets....
$endgroup$
– user52269
Sep 8 '18 at 6:57
3
3
$begingroup$
I like this abuse of javascript a lot. !0 for true, +!0 for 1, [1] for array containing 1, [1]+[0] for concatenate two arrays to get the string "10" ; subtract 1 from that twice to end up as 8. Love it.
$endgroup$
– Ross Presser
Sep 8 '18 at 4:10
$begingroup$
I like this abuse of javascript a lot. !0 for true, +!0 for 1, [1] for array containing 1, [1]+[0] for concatenate two arrays to get the string "10" ; subtract 1 from that twice to end up as 8. Love it.
$endgroup$
– Ross Presser
Sep 8 '18 at 4:10
2
2
$begingroup$
but the question is tagged mathematics and not programming
$endgroup$
– phuclv
Sep 8 '18 at 4:21
$begingroup$
but the question is tagged mathematics and not programming
$endgroup$
– phuclv
Sep 8 '18 at 4:21
3
3
$begingroup$
Who says you can use square brackets....
$endgroup$
– user52269
Sep 8 '18 at 6:57
$begingroup$
Who says you can use square brackets....
$endgroup$
– user52269
Sep 8 '18 at 6:57
add a comment |
8
$begingroup$
$concat(0!,0) - 0! - 0! = 8$
becomes:
$concat(1,0) - 1 - 1 = 8$
and finally:
$10 - 2 = 8$
cool puzzle!
answered Sep 7 '18 at 5:38
ArnaldurArnaldur
971
$endgroup$
2
$begingroup$
Great answer! But it has already been answered (check out @malioboro 's answer). Nonetheless, welcome to the Puzzling Stack Exchange (Puzzling.SE)! Since you are new to the Stack Exchange community, let alone this site, I strongly suggest that you visit the Help Center for more info; in particular, I suggest going here, then here for questions (not answers) :D
$endgroup$
– user477343
Sep 7 '18 at 7:44
add a comment |
8
$begingroup$
$concat(0!,0) - 0! - 0! = 8$
becomes:
$concat(1,0) - 1 - 1 = 8$
and finally:
$10 - 2 = 8$
cool puzzle!
answered Sep 7 '18 at 5:38
ArnaldurArnaldur
971
$endgroup$
2
$begingroup$
Great answer! But it has already been answered (check out @malioboro 's answer). Nonetheless, welcome to the Puzzling Stack Exchange (Puzzling.SE)! Since you are new to the Stack Exchange community, let alone this site, I strongly suggest that you visit the Help Center for more info; in particular, I suggest going here, then here for questions (not answers) :D
$endgroup$
– user477343
Sep 7 '18 at 7:44
add a comment |
8
8
8
$begingroup$
$concat(0!,0) - 0! - 0! = 8$
becomes:
$concat(1,0) - 1 - 1 = 8$
and finally:
$10 - 2 = 8$
cool puzzle!
answered Sep 7 '18 at 5:38
ArnaldurArnaldur
971
$endgroup$
$concat(0!,0) - 0! - 0! = 8$
becomes:
$concat(1,0) - 1 - 1 = 8$
and finally:
$10 - 2 = 8$
cool puzzle!
answered Sep 7 '18 at 5:38
ArnaldurArnaldur
971
answered Sep 7 '18 at 5:38
ArnaldurArnaldur
971
answered Sep 7 '18 at 5:38
ArnaldurArnaldur
971
answered Sep 7 '18 at 5:38
ArnaldurArnaldur
971
971
2
$begingroup$
Great answer! But it has already been answered (check out @malioboro 's answer). Nonetheless, welcome to the Puzzling Stack Exchange (Puzzling.SE)! Since you are new to the Stack Exchange community, let alone this site, I strongly suggest that you visit the Help Center for more info; in particular, I suggest going here, then here for questions (not answers) :D
$endgroup$
– user477343
Sep 7 '18 at 7:44
add a comment |
2
$begingroup$
Great answer! But it has already been answered (check out @malioboro 's answer). Nonetheless, welcome to the Puzzling Stack Exchange (Puzzling.SE)! Since you are new to the Stack Exchange community, let alone this site, I strongly suggest that you visit the Help Center for more info; in particular, I suggest going here, then here for questions (not answers) :D
$endgroup$
– user477343
Sep 7 '18 at 7:44
2
2
$begingroup$
Great answer! But it has already been answered (check out @malioboro 's answer). Nonetheless, welcome to the Puzzling Stack Exchange (Puzzling.SE)! Since you are new to the Stack Exchange community, let alone this site, I strongly suggest that you visit the Help Center for more info; in particular, I suggest going here, then here for questions (not answers) :D
$endgroup$
– user477343
Sep 7 '18 at 7:44
$begingroup$
Great answer! But it has already been answered (check out @malioboro 's answer). Nonetheless, welcome to the Puzzling Stack Exchange (Puzzling.SE)! Since you are new to the Stack Exchange community, let alone this site, I strongly suggest that you visit the Help Center for more info; in particular, I suggest going here, then here for questions (not answers) :D
$endgroup$
– user477343
Sep 7 '18 at 7:44
add a comment |
7
$begingroup$
$0 + 0 + 0 + 0 equiv 8$
Adding the symbol $-$ over the equals sign makes it a congruence sign.
Considering the congruence relation, we must be working mod N, where N divides 8.
answered Sep 8 '18 at 6:39
VaelusVaelus
40124
$endgroup$
add a comment |
7
$begingroup$
$0 + 0 + 0 + 0 equiv 8$
Adding the symbol $-$ over the equals sign makes it a congruence sign.
Considering the congruence relation, we must be working mod N, where N divides 8.
answered Sep 8 '18 at 6:39
VaelusVaelus
40124
$endgroup$
add a comment |
7
7
7
$begingroup$
$0 + 0 + 0 + 0 equiv 8$
Adding the symbol $-$ over the equals sign makes it a congruence sign.
Considering the congruence relation, we must be working mod N, where N divides 8.
answered Sep 8 '18 at 6:39
VaelusVaelus
40124
$endgroup$
$0 + 0 + 0 + 0 equiv 8$
Adding the symbol $-$ over the equals sign makes it a congruence sign.
Considering the congruence relation, we must be working mod N, where N divides 8.
answered Sep 8 '18 at 6:39
VaelusVaelus
40124
answered Sep 8 '18 at 6:39
VaelusVaelus
40124
answered Sep 8 '18 at 6:39
VaelusVaelus
40124
answered Sep 8 '18 at 6:39
VaelusVaelus
40124
40124
add a comment |
add a comment |
3
$begingroup$
Just put the minus symbol over the first zero to give it the look of a figure 8 and use plus to add the zeros.
answered Sep 7 '18 at 11:37
SentinelSentinel
1,041212
$endgroup$
$begingroup$
Great answer, but unfortunately, this is a duplicate of @TheSimpliFire 's answer.
$endgroup$
– user477343
Sep 7 '18 at 11:51
2
$begingroup$
@user477343 No, not quite, since this answer puts a line through the 0 instead of putting two 0s on top of each other. In this answer, only one 0 becomes an 8.
$endgroup$
– Fabian Röling
Sep 7 '18 at 12:02
$begingroup$
@FabianRöling As I read it, it uses the minus symbol to be put over the first zero and make it look like an $8$. In my imagination, I assumed the zero is the bottom circle of $8$ since there is something over it, of which is another circle $largecirc$. I did not fully understand what this answer was talking about in saying to use a subtraction symbol, but just assumed it was just like TheSimpliFire's answer due to their similarity; and since that is a great answer, I thought this was great, too. If you understand this answer, may you please explain to me in other words if you can? :
$endgroup$
– user477343
Sep 7 '18 at 12:09
3
$begingroup$
@user477343 It uses the horizontal line of the minus sign to cut through the middle of the $0$. It is "over" the zero in the sense of overwritten.
$endgroup$
– Jaap Scherphuis
Sep 7 '18 at 12:55
1
$begingroup$
Yeah. 'over' in the sense of "in front of" on a dimension between you and your monitor.
$endgroup$
– Sentinel
Sep 7 '18 at 14:44
|
show 2 more comments
3
$begingroup$
Just put the minus symbol over the first zero to give it the look of a figure 8 and use plus to add the zeros.
answered Sep 7 '18 at 11:37
SentinelSentinel
1,041212
$endgroup$
$begingroup$
Great answer, but unfortunately, this is a duplicate of @TheSimpliFire 's answer.
$endgroup$
– user477343
Sep 7 '18 at 11:51
2
$begingroup$
@user477343 No, not quite, since this answer puts a line through the 0 instead of putting two 0s on top of each other. In this answer, only one 0 becomes an 8.
$endgroup$
– Fabian Röling
Sep 7 '18 at 12:02
$begingroup$
@FabianRöling As I read it, it uses the minus symbol to be put over the first zero and make it look like an $8$. In my imagination, I assumed the zero is the bottom circle of $8$ since there is something over it, of which is another circle $largecirc$. I did not fully understand what this answer was talking about in saying to use a subtraction symbol, but just assumed it was just like TheSimpliFire's answer due to their similarity; and since that is a great answer, I thought this was great, too. If you understand this answer, may you please explain to me in other words if you can? :
$endgroup$
– user477343
Sep 7 '18 at 12:09
3
$begingroup$
@user477343 It uses the horizontal line of the minus sign to cut through the middle of the $0$. It is "over" the zero in the sense of overwritten.
$endgroup$
– Jaap Scherphuis
Sep 7 '18 at 12:55
1
$begingroup$
Yeah. 'over' in the sense of "in front of" on a dimension between you and your monitor.
$endgroup$
– Sentinel
Sep 7 '18 at 14:44
|
show 2 more comments
3
3
3
$begingroup$
Just put the minus symbol over the first zero to give it the look of a figure 8 and use plus to add the zeros.
answered Sep 7 '18 at 11:37
SentinelSentinel
1,041212
$endgroup$
Just put the minus symbol over the first zero to give it the look of a figure 8 and use plus to add the zeros.
answered Sep 7 '18 at 11:37
SentinelSentinel
1,041212
answered Sep 7 '18 at 11:37
SentinelSentinel
1,041212
answered Sep 7 '18 at 11:37
SentinelSentinel
1,041212
answered Sep 7 '18 at 11:37
SentinelSentinel
1,041212
1,041212
$begingroup$
Great answer, but unfortunately, this is a duplicate of @TheSimpliFire 's answer.
$endgroup$
– user477343
Sep 7 '18 at 11:51
2
$begingroup$
@user477343 No, not quite, since this answer puts a line through the 0 instead of putting two 0s on top of each other. In this answer, only one 0 becomes an 8.
$endgroup$
– Fabian Röling
Sep 7 '18 at 12:02
$begingroup$
@FabianRöling As I read it, it uses the minus symbol to be put over the first zero and make it look like an $8$. In my imagination, I assumed the zero is the bottom circle of $8$ since there is something over it, of which is another circle $largecirc$. I did not fully understand what this answer was talking about in saying to use a subtraction symbol, but just assumed it was just like TheSimpliFire's answer due to their similarity; and since that is a great answer, I thought this was great, too. If you understand this answer, may you please explain to me in other words if you can? :
$endgroup$
– user477343
Sep 7 '18 at 12:09
3
$begingroup$
@user477343 It uses the horizontal line of the minus sign to cut through the middle of the $0$. It is "over" the zero in the sense of overwritten.
$endgroup$
– Jaap Scherphuis
Sep 7 '18 at 12:55
1
$begingroup$
Yeah. 'over' in the sense of "in front of" on a dimension between you and your monitor.
$endgroup$
– Sentinel
Sep 7 '18 at 14:44
|
show 2 more comments
$begingroup$
Great answer, but unfortunately, this is a duplicate of @TheSimpliFire 's answer.
$endgroup$
– user477343
Sep 7 '18 at 11:51
2
$begingroup$
@user477343 No, not quite, since this answer puts a line through the 0 instead of putting two 0s on top of each other. In this answer, only one 0 becomes an 8.
$endgroup$
– Fabian Röling
Sep 7 '18 at 12:02
$begingroup$
@FabianRöling As I read it, it uses the minus symbol to be put over the first zero and make it look like an $8$. In my imagination, I assumed the zero is the bottom circle of $8$ since there is something over it, of which is another circle $largecirc$. I did not fully understand what this answer was talking about in saying to use a subtraction symbol, but just assumed it was just like TheSimpliFire's answer due to their similarity; and since that is a great answer, I thought this was great, too. If you understand this answer, may you please explain to me in other words if you can? :
$endgroup$
– user477343
Sep 7 '18 at 12:09
3
$begingroup$
@user477343 It uses the horizontal line of the minus sign to cut through the middle of the $0$. It is "over" the zero in the sense of overwritten.
$endgroup$
– Jaap Scherphuis
Sep 7 '18 at 12:55
1
$begingroup$
Yeah. 'over' in the sense of "in front of" on a dimension between you and your monitor.
$endgroup$
– Sentinel
Sep 7 '18 at 14:44
$begingroup$
Great answer, but unfortunately, this is a duplicate of @TheSimpliFire 's answer.
$endgroup$
– user477343
Sep 7 '18 at 11:51
$begingroup$
Great answer, but unfortunately, this is a duplicate of @TheSimpliFire 's answer.
$endgroup$
– user477343
Sep 7 '18 at 11:51
2
2
$begingroup$
@user477343 No, not quite, since this answer puts a line through the 0 instead of putting two 0s on top of each other. In this answer, only one 0 becomes an 8.
$endgroup$
– Fabian Röling
Sep 7 '18 at 12:02
$begingroup$
@user477343 No, not quite, since this answer puts a line through the 0 instead of putting two 0s on top of each other. In this answer, only one 0 becomes an 8.
$endgroup$
– Fabian Röling
Sep 7 '18 at 12:02
$begingroup$
@FabianRöling As I read it, it uses the minus symbol to be put over the first zero and make it look like an $8$. In my imagination, I assumed the zero is the bottom circle of $8$ since there is something over it, of which is another circle $largecirc$. I did not fully understand what this answer was talking about in saying to use a subtraction symbol, but just assumed it was just like TheSimpliFire's answer due to their similarity; and since that is a great answer, I thought this was great, too. If you understand this answer, may you please explain to me in other words if you can? :
$endgroup$
– user477343
Sep 7 '18 at 12:09
$begingroup$
@FabianRöling As I read it, it uses the minus symbol to be put over the first zero and make it look like an $8$. In my imagination, I assumed the zero is the bottom circle of $8$ since there is something over it, of which is another circle $largecirc$. I did not fully understand what this answer was talking about in saying to use a subtraction symbol, but just assumed it was just like TheSimpliFire's answer due to their similarity; and since that is a great answer, I thought this was great, too. If you understand this answer, may you please explain to me in other words if you can? :
$endgroup$
– user477343
Sep 7 '18 at 12:09
3
3
$begingroup$
@user477343 It uses the horizontal line of the minus sign to cut through the middle of the $0$. It is "over" the zero in the sense of overwritten.
$endgroup$
– Jaap Scherphuis
Sep 7 '18 at 12:55
$begingroup$
@user477343 It uses the horizontal line of the minus sign to cut through the middle of the $0$. It is "over" the zero in the sense of overwritten.
$endgroup$
– Jaap Scherphuis
Sep 7 '18 at 12:55
1
1
$begingroup$
Yeah. 'over' in the sense of "in front of" on a dimension between you and your monitor.
$endgroup$
– Sentinel
Sep 7 '18 at 14:44
$begingroup$
Yeah. 'over' in the sense of "in front of" on a dimension between you and your monitor.
$endgroup$
– Sentinel
Sep 7 '18 at 14:44
|
show 2 more comments
3
$begingroup$
Here is an answer that doesn't use the semi-factorial or any concatenation.
$$0 + 0 - 0! / 0 = (-8)!$$
The left side is $-1/0$ and the right side is $-infty$.
Plugging the expression into Wolfram.
answered Sep 8 '18 at 0:20
user1717828user1717828
1,514617
$endgroup$
4
$begingroup$
Division by zero is undefined (the limit of 0!/x as x goes to 0 is plus or minus infinity) and so are negative integer factorials (they are not equivalent to negative infinity). Even if they both evaluated to +∞, you can prove nonsense assuming you can equate infinities and treat like a real number. E.g., |1/0| = +∞, adding 1 to positive infinity is still positive infinity, hence |1/0| + 1 = +∞, hence |1/0| = |1/0| +1 by transitivity of equality. Subtract |1/0| from both sides and prove 0 = 1.
$endgroup$
– dr jimbob
Sep 8 '18 at 18:11
add a comment |
3
$begingroup$
Here is an answer that doesn't use the semi-factorial or any concatenation.
$$0 + 0 - 0! / 0 = (-8)!$$
The left side is $-1/0$ and the right side is $-infty$.
Plugging the expression into Wolfram.
answered Sep 8 '18 at 0:20
user1717828user1717828
1,514617
$endgroup$
4
$begingroup$
Division by zero is undefined (the limit of 0!/x as x goes to 0 is plus or minus infinity) and so are negative integer factorials (they are not equivalent to negative infinity). Even if they both evaluated to +∞, you can prove nonsense assuming you can equate infinities and treat like a real number. E.g., |1/0| = +∞, adding 1 to positive infinity is still positive infinity, hence |1/0| + 1 = +∞, hence |1/0| = |1/0| +1 by transitivity of equality. Subtract |1/0| from both sides and prove 0 = 1.
$endgroup$
– dr jimbob
Sep 8 '18 at 18:11
add a comment |
3
3
3
$begingroup$
Here is an answer that doesn't use the semi-factorial or any concatenation.
$$0 + 0 - 0! / 0 = (-8)!$$
The left side is $-1/0$ and the right side is $-infty$.
Plugging the expression into Wolfram.
answered Sep 8 '18 at 0:20
user1717828user1717828
1,514617
$endgroup$
Here is an answer that doesn't use the semi-factorial or any concatenation.
$$0 + 0 - 0! / 0 = (-8)!$$
The left side is $-1/0$ and the right side is $-infty$.
Plugging the expression into Wolfram.
answered Sep 8 '18 at 0:20
user1717828user1717828
1,514617
answered Sep 8 '18 at 0:20
user1717828user1717828
1,514617
answered Sep 8 '18 at 0:20
user1717828user1717828
1,514617
answered Sep 8 '18 at 0:20
user1717828user1717828
1,514617
1,514617
4
$begingroup$
Division by zero is undefined (the limit of 0!/x as x goes to 0 is plus or minus infinity) and so are negative integer factorials (they are not equivalent to negative infinity). Even if they both evaluated to +∞, you can prove nonsense assuming you can equate infinities and treat like a real number. E.g., |1/0| = +∞, adding 1 to positive infinity is still positive infinity, hence |1/0| + 1 = +∞, hence |1/0| = |1/0| +1 by transitivity of equality. Subtract |1/0| from both sides and prove 0 = 1.
$endgroup$
– dr jimbob
Sep 8 '18 at 18:11
add a comment |
4
$begingroup$
Division by zero is undefined (the limit of 0!/x as x goes to 0 is plus or minus infinity) and so are negative integer factorials (they are not equivalent to negative infinity). Even if they both evaluated to +∞, you can prove nonsense assuming you can equate infinities and treat like a real number. E.g., |1/0| = +∞, adding 1 to positive infinity is still positive infinity, hence |1/0| + 1 = +∞, hence |1/0| = |1/0| +1 by transitivity of equality. Subtract |1/0| from both sides and prove 0 = 1.
$endgroup$
– dr jimbob
Sep 8 '18 at 18:11
4
4
$begingroup$
Division by zero is undefined (the limit of 0!/x as x goes to 0 is plus or minus infinity) and so are negative integer factorials (they are not equivalent to negative infinity). Even if they both evaluated to +∞, you can prove nonsense assuming you can equate infinities and treat like a real number. E.g., |1/0| = +∞, adding 1 to positive infinity is still positive infinity, hence |1/0| + 1 = +∞, hence |1/0| = |1/0| +1 by transitivity of equality. Subtract |1/0| from both sides and prove 0 = 1.
$endgroup$
– dr jimbob
Sep 8 '18 at 18:11
$begingroup$
Division by zero is undefined (the limit of 0!/x as x goes to 0 is plus or minus infinity) and so are negative integer factorials (they are not equivalent to negative infinity). Even if they both evaluated to +∞, you can prove nonsense assuming you can equate infinities and treat like a real number. E.g., |1/0| = +∞, adding 1 to positive infinity is still positive infinity, hence |1/0| + 1 = +∞, hence |1/0| = |1/0| +1 by transitivity of equality. Subtract |1/0| from both sides and prove 0 = 1.
$endgroup$
– dr jimbob
Sep 8 '18 at 18:11
add a comment |
3
$begingroup$
concatenate(0!, 0) - concatenate(0! + 0!) = 8.
Note that 0! = 1
(0!, 0) = 10, and (0! + 0!) = 2, so 10 - 2 = 8
edited Sep 11 '18 at 1:59
Alto
1,222320
answered Sep 11 '18 at 0:57
NickNick
291
$endgroup$
$begingroup$
Unfortunately concatenate is not a valid operation.
$endgroup$
– boboquack
Sep 11 '18 at 1:37
$begingroup$
Congrats, concatenation is allowed now! Good answer, but people beat you to it. Welcome to Puzzling.SE! Here's a bonus puzzle: What's the word? Hilarious - extremely amusing
$endgroup$
– Alto
Sep 11 '18 at 1:47
$begingroup$
Also, concatenation is combination. Basically, (2, 4) = 24, you get it.
$endgroup$
– Alto
Sep 11 '18 at 1:48
add a comment |
3
$begingroup$
concatenate(0!, 0) - concatenate(0! + 0!) = 8.
Note that 0! = 1
(0!, 0) = 10, and (0! + 0!) = 2, so 10 - 2 = 8
edited Sep 11 '18 at 1:59
Alto
1,222320
answered Sep 11 '18 at 0:57
NickNick
291
$endgroup$
$begingroup$
Unfortunately concatenate is not a valid operation.
$endgroup$
– boboquack
Sep 11 '18 at 1:37
$begingroup$
Congrats, concatenation is allowed now! Good answer, but people beat you to it. Welcome to Puzzling.SE! Here's a bonus puzzle: What's the word? Hilarious - extremely amusing
$endgroup$
– Alto
Sep 11 '18 at 1:47
$begingroup$
Also, concatenation is combination. Basically, (2, 4) = 24, you get it.
$endgroup$
– Alto
Sep 11 '18 at 1:48
add a comment |
3
3
3
$begingroup$
concatenate(0!, 0) - concatenate(0! + 0!) = 8.
Note that 0! = 1
(0!, 0) = 10, and (0! + 0!) = 2, so 10 - 2 = 8
edited Sep 11 '18 at 1:59
Alto
1,222320
answered Sep 11 '18 at 0:57
NickNick
291
$endgroup$
concatenate(0!, 0) - concatenate(0! + 0!) = 8.
Note that 0! = 1
(0!, 0) = 10, and (0! + 0!) = 2, so 10 - 2 = 8
edited Sep 11 '18 at 1:59
Alto
1,222320
answered Sep 11 '18 at 0:57
NickNick
291
edited Sep 11 '18 at 1:59
Alto
1,222320
edited Sep 11 '18 at 1:59
Alto
1,222320
edited Sep 11 '18 at 1:59
Alto
1,222320
1,222320
answered Sep 11 '18 at 0:57
NickNick
291
answered Sep 11 '18 at 0:57
NickNick
291
answered Sep 11 '18 at 0:57
NickNick
291
291
$begingroup$
Unfortunately concatenate is not a valid operation.
$endgroup$
– boboquack
Sep 11 '18 at 1:37
$begingroup$
Congrats, concatenation is allowed now! Good answer, but people beat you to it. Welcome to Puzzling.SE! Here's a bonus puzzle: What's the word? Hilarious - extremely amusing
$endgroup$
– Alto
Sep 11 '18 at 1:47
$begingroup$
Also, concatenation is combination. Basically, (2, 4) = 24, you get it.
$endgroup$
– Alto
Sep 11 '18 at 1:48
add a comment |
$begingroup$
Unfortunately concatenate is not a valid operation.
$endgroup$
– boboquack
Sep 11 '18 at 1:37
$begingroup$
Congrats, concatenation is allowed now! Good answer, but people beat you to it. Welcome to Puzzling.SE! Here's a bonus puzzle: What's the word? Hilarious - extremely amusing
$endgroup$
– Alto
Sep 11 '18 at 1:47
$begingroup$
Also, concatenation is combination. Basically, (2, 4) = 24, you get it.
$endgroup$
– Alto
Sep 11 '18 at 1:48
$begingroup$
Unfortunately concatenate is not a valid operation.
$endgroup$
– boboquack
Sep 11 '18 at 1:37
$begingroup$
Unfortunately concatenate is not a valid operation.
$endgroup$
– boboquack
Sep 11 '18 at 1:37
$begingroup$
Congrats, concatenation is allowed now! Good answer, but people beat you to it. Welcome to Puzzling.SE! Here's a bonus puzzle: What's the word? Hilarious - extremely amusing
$endgroup$
– Alto
Sep 11 '18 at 1:47
$begingroup$
Congrats, concatenation is allowed now! Good answer, but people beat you to it. Welcome to Puzzling.SE! Here's a bonus puzzle: What's the word? Hilarious - extremely amusing
$endgroup$
– Alto
Sep 11 '18 at 1:47
$begingroup$
Also, concatenation is combination. Basically, (2, 4) = 24, you get it.
$endgroup$
– Alto
Sep 11 '18 at 1:48
$begingroup$
Also, concatenation is combination. Basically, (2, 4) = 24, you get it.
$endgroup$
– Alto
Sep 11 '18 at 1:48
add a comment |
3
$begingroup$
This could work too:
(0!+0+0)/0 = ∞
Explanation
(0!+0+0)/0 = 1/0 which is infinity (8 but put horizontally)
answered Sep 13 '18 at 9:10
Michal B.Michal B.
2665
$endgroup$
$begingroup$
That is not infinity — it is undefined :
$endgroup$
– user477343
Oct 3 '18 at 11:33
add a comment |
3
$begingroup$
This could work too:
(0!+0+0)/0 = ∞
Explanation
(0!+0+0)/0 = 1/0 which is infinity (8 but put horizontally)
answered Sep 13 '18 at 9:10
Michal B.Michal B.
2665
$endgroup$
$begingroup$
That is not infinity — it is undefined :
$endgroup$
– user477343
Oct 3 '18 at 11:33
add a comment |
3
3
3
$begingroup$
This could work too:
(0!+0+0)/0 = ∞
Explanation
(0!+0+0)/0 = 1/0 which is infinity (8 but put horizontally)
answered Sep 13 '18 at 9:10
Michal B.Michal B.
2665
$endgroup$
This could work too:
(0!+0+0)/0 = ∞
Explanation
(0!+0+0)/0 = 1/0 which is infinity (8 but put horizontally)
answered Sep 13 '18 at 9:10
Michal B.Michal B.
2665
answered Sep 13 '18 at 9:10
Michal B.Michal B.
2665
answered Sep 13 '18 at 9:10
Michal B.Michal B.
2665
answered Sep 13 '18 at 9:10
Michal B.Michal B.
2665
2665
$begingroup$
That is not infinity — it is undefined :
$endgroup$
– user477343
Oct 3 '18 at 11:33
add a comment |
$begingroup$
That is not infinity — it is undefined :
$endgroup$
– user477343
Oct 3 '18 at 11:33
$begingroup$
That is not infinity — it is undefined :
$endgroup$
– user477343
Oct 3 '18 at 11:33
$begingroup$
That is not infinity — it is undefined :
$endgroup$
– user477343
Oct 3 '18 at 11:33
add a comment |
3
$begingroup$
Add a - above the equals to get $0000 equiv 8$, which is true assuming we are working in the ring $mathbb{Z}/mathbb{2Z}$. (Note I'm trying to avoid writing $[0] = [8]$...)
edited Sep 14 '18 at 2:45
a stone arachnid
1697
answered Sep 11 '18 at 15:52
Gus314Gus314
312
$endgroup$
add a comment |
3
$begingroup$
Add a - above the equals to get $0000 equiv 8$, which is true assuming we are working in the ring $mathbb{Z}/mathbb{2Z}$. (Note I'm trying to avoid writing $[0] = [8]$...)
edited Sep 14 '18 at 2:45
a stone arachnid
1697
answered Sep 11 '18 at 15:52
Gus314Gus314
312
$endgroup$
add a comment |
3
3
3
$begingroup$
Add a - above the equals to get $0000 equiv 8$, which is true assuming we are working in the ring $mathbb{Z}/mathbb{2Z}$. (Note I'm trying to avoid writing $[0] = [8]$...)
edited Sep 14 '18 at 2:45
a stone arachnid
1697
answered Sep 11 '18 at 15:52
Gus314Gus314
312
$endgroup$
Add a - above the equals to get $0000 equiv 8$, which is true assuming we are working in the ring $mathbb{Z}/mathbb{2Z}$. (Note I'm trying to avoid writing $[0] = [8]$...)
edited Sep 14 '18 at 2:45
a stone arachnid
1697
answered Sep 11 '18 at 15:52
Gus314Gus314
312
edited Sep 14 '18 at 2:45
a stone arachnid
1697
edited Sep 14 '18 at 2:45
a stone arachnid
1697
edited Sep 14 '18 at 2:45
a stone arachnid
1697
1697
answered Sep 11 '18 at 15:52
Gus314Gus314
312
answered Sep 11 '18 at 15:52
Gus314Gus314
312
answered Sep 11 '18 at 15:52
Gus314Gus314
312
312
add a comment |
add a comment |
2
$begingroup$
Question limits the symbols, not the operations. So with the symbol + can make the operator ++.
(++(++(++(++(++(++(++0000))))))) = 8
answered Sep 7 '18 at 21:23
David Browne - MicrosoftDavid Browne - Microsoft
1294
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$begingroup$
I think you'd need the prefix form if you actually wanted that to work.
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– LegionMammal978
Sep 7 '18 at 23:48
1
$begingroup$
Doh. Of course you can't increment a literal in the first place.
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– David Browne - Microsoft
Sep 7 '18 at 23:55
1
$begingroup$
but that symbol doesn't exist in mathematics
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– phuclv
Sep 8 '18 at 4:20
2
$begingroup$
You need to pass an l-value to the++and--operators, and neither their result or a literal is one.
$endgroup$
– NieDzejkob
Sep 8 '18 at 14:35
add a comment |
2
$begingroup$
Question limits the symbols, not the operations. So with the symbol + can make the operator ++.
(++(++(++(++(++(++(++0000))))))) = 8
answered Sep 7 '18 at 21:23
David Browne - MicrosoftDavid Browne - Microsoft
1294
$endgroup$
$begingroup$
I think you'd need the prefix form if you actually wanted that to work.
$endgroup$
– LegionMammal978
Sep 7 '18 at 23:48
1
$begingroup$
Doh. Of course you can't increment a literal in the first place.
$endgroup$
– David Browne - Microsoft
Sep 7 '18 at 23:55
1
$begingroup$
but that symbol doesn't exist in mathematics
$endgroup$
– phuclv
Sep 8 '18 at 4:20
2
$begingroup$
You need to pass an l-value to the++and--operators, and neither their result or a literal is one.
$endgroup$
– NieDzejkob
Sep 8 '18 at 14:35
add a comment |
2
2
2
$begingroup$
Question limits the symbols, not the operations. So with the symbol + can make the operator ++.
(++(++(++(++(++(++(++0000))))))) = 8
answered Sep 7 '18 at 21:23
David Browne - MicrosoftDavid Browne - Microsoft
1294
$endgroup$
Question limits the symbols, not the operations. So with the symbol + can make the operator ++.
(++(++(++(++(++(++(++0000))))))) = 8
answered Sep 7 '18 at 21:23
David Browne - MicrosoftDavid Browne - Microsoft
1294
edited Sep 7 '18 at 23:53
answered Sep 7 '18 at 21:23
David Browne - MicrosoftDavid Browne - Microsoft
1294
answered Sep 7 '18 at 21:23
David Browne - MicrosoftDavid Browne - Microsoft
1294
answered Sep 7 '18 at 21:23
David Browne - MicrosoftDavid Browne - Microsoft
1294
1294
$begingroup$
I think you'd need the prefix form if you actually wanted that to work.
$endgroup$
– LegionMammal978
Sep 7 '18 at 23:48
1
$begingroup$
Doh. Of course you can't increment a literal in the first place.
$endgroup$
– David Browne - Microsoft
Sep 7 '18 at 23:55
1
$begingroup$
but that symbol doesn't exist in mathematics
$endgroup$
– phuclv
Sep 8 '18 at 4:20
2
$begingroup$
You need to pass an l-value to the++and--operators, and neither their result or a literal is one.
$endgroup$
– NieDzejkob
Sep 8 '18 at 14:35
add a comment |
$begingroup$
I think you'd need the prefix form if you actually wanted that to work.
$endgroup$
– LegionMammal978
Sep 7 '18 at 23:48
1
$begingroup$
Doh. Of course you can't increment a literal in the first place.
$endgroup$
– David Browne - Microsoft
Sep 7 '18 at 23:55
1
$begingroup$
but that symbol doesn't exist in mathematics
$endgroup$
– phuclv
Sep 8 '18 at 4:20
2
$begingroup$
You need to pass an l-value to the++and--operators, and neither their result or a literal is one.
$endgroup$
– NieDzejkob
Sep 8 '18 at 14:35
$begingroup$
I think you'd need the prefix form if you actually wanted that to work.
$endgroup$
– LegionMammal978
Sep 7 '18 at 23:48
$begingroup$
I think you'd need the prefix form if you actually wanted that to work.
$endgroup$
– LegionMammal978
Sep 7 '18 at 23:48
1
1
$begingroup$
Doh. Of course you can't increment a literal in the first place.
$endgroup$
– David Browne - Microsoft
Sep 7 '18 at 23:55
$begingroup$
Doh. Of course you can't increment a literal in the first place.
$endgroup$
– David Browne - Microsoft
Sep 7 '18 at 23:55
1
1
$begingroup$
but that symbol doesn't exist in mathematics
$endgroup$
– phuclv
Sep 8 '18 at 4:20
$begingroup$
but that symbol doesn't exist in mathematics
$endgroup$
– phuclv
Sep 8 '18 at 4:20
2
2
$begingroup$
You need to pass an l-value to the
++ and -- operators, and neither their result or a literal is one.$endgroup$
– NieDzejkob
Sep 8 '18 at 14:35
$begingroup$
You need to pass an l-value to the
++ and -- operators, and neither their result or a literal is one.$endgroup$
– NieDzejkob
Sep 8 '18 at 14:35
add a comment |
0
$begingroup$
$ ((0! + 0!)$)*(0! + 0!) = 2^2*2 = 8 $.
Further explanation:
The $$$ operation denotes the superfactorial defined as : $ n$ = (n!) uparrow uparrow (n!)$.
answered Sep 7 '18 at 14:28
RiaRia
354111
$endgroup$
3
$begingroup$
Good idea, but make sure to read the question fully. It gives you a list of what operations can be used and $ is not one of them.
$endgroup$
– Sensoray
Sep 7 '18 at 15:26
add a comment |
0
$begingroup$
$ ((0! + 0!)$)*(0! + 0!) = 2^2*2 = 8 $.
Further explanation:
The $$$ operation denotes the superfactorial defined as : $ n$ = (n!) uparrow uparrow (n!)$.
answered Sep 7 '18 at 14:28
RiaRia
354111
$endgroup$
3
$begingroup$
Good idea, but make sure to read the question fully. It gives you a list of what operations can be used and $ is not one of them.
$endgroup$
– Sensoray
Sep 7 '18 at 15:26
add a comment |
0
0
0
$begingroup$
$ ((0! + 0!)$)*(0! + 0!) = 2^2*2 = 8 $.
Further explanation:
The $$$ operation denotes the superfactorial defined as : $ n$ = (n!) uparrow uparrow (n!)$.
answered Sep 7 '18 at 14:28
RiaRia
354111
$endgroup$
$ ((0! + 0!)$)*(0! + 0!) = 2^2*2 = 8 $.
Further explanation:
The $$$ operation denotes the superfactorial defined as : $ n$ = (n!) uparrow uparrow (n!)$.
answered Sep 7 '18 at 14:28
RiaRia
354111
edited Sep 7 '18 at 14:41
answered Sep 7 '18 at 14:28
RiaRia
354111
answered Sep 7 '18 at 14:28
RiaRia
354111
answered Sep 7 '18 at 14:28
RiaRia
354111
354111
3
$begingroup$
Good idea, but make sure to read the question fully. It gives you a list of what operations can be used and $ is not one of them.
$endgroup$
– Sensoray
Sep 7 '18 at 15:26
add a comment |
3
$begingroup$
Good idea, but make sure to read the question fully. It gives you a list of what operations can be used and $ is not one of them.
$endgroup$
– Sensoray
Sep 7 '18 at 15:26
3
3
$begingroup$
Good idea, but make sure to read the question fully. It gives you a list of what operations can be used and $ is not one of them.
$endgroup$
– Sensoray
Sep 7 '18 at 15:26
$begingroup$
Good idea, but make sure to read the question fully. It gives you a list of what operations can be used and $ is not one of them.
$endgroup$
– Sensoray
Sep 7 '18 at 15:26
add a comment |
0
$begingroup$
If you turn the problem around
answered Sep 8 '18 at 8:09
Florian FFlorian F
9,22612260
$endgroup$
$begingroup$
Similar to @rrauenza's answer
$endgroup$
– TheSimpliFire
Sep 8 '18 at 8:11
$begingroup$
Yes. A different way to express the same idea.
$endgroup$
– Florian F
Sep 8 '18 at 9:13
add a comment |
0
$begingroup$
If you turn the problem around
answered Sep 8 '18 at 8:09
Florian FFlorian F
9,22612260
$endgroup$
$begingroup$
Similar to @rrauenza's answer
$endgroup$
– TheSimpliFire
Sep 8 '18 at 8:11
$begingroup$
Yes. A different way to express the same idea.
$endgroup$
– Florian F
Sep 8 '18 at 9:13
add a comment |
0
0
0
$begingroup$
If you turn the problem around
answered Sep 8 '18 at 8:09
Florian FFlorian F
9,22612260
$endgroup$
If you turn the problem around
answered Sep 8 '18 at 8:09
Florian FFlorian F
9,22612260
answered Sep 8 '18 at 8:09
Florian FFlorian F
9,22612260
answered Sep 8 '18 at 8:09
Florian FFlorian F
9,22612260
answered Sep 8 '18 at 8:09
Florian FFlorian F
9,22612260
9,22612260
$begingroup$
Similar to @rrauenza's answer
$endgroup$
– TheSimpliFire
Sep 8 '18 at 8:11
$begingroup$
Yes. A different way to express the same idea.
$endgroup$
– Florian F
Sep 8 '18 at 9:13
add a comment |
$begingroup$
Similar to @rrauenza's answer
$endgroup$
– TheSimpliFire
Sep 8 '18 at 8:11
$begingroup$
Yes. A different way to express the same idea.
$endgroup$
– Florian F
Sep 8 '18 at 9:13
$begingroup$
Similar to @rrauenza's answer
$endgroup$
– TheSimpliFire
Sep 8 '18 at 8:11
$begingroup$
Similar to @rrauenza's answer
$endgroup$
– TheSimpliFire
Sep 8 '18 at 8:11
$begingroup$
Yes. A different way to express the same idea.
$endgroup$
– Florian F
Sep 8 '18 at 9:13
$begingroup$
Yes. A different way to express the same idea.
$endgroup$
– Florian F
Sep 8 '18 at 9:13
add a comment |
0
$begingroup$
Similar to @Vaelus
$0+0+0+0 leq 8$
Explanation
You can get the $leq$ by adding a $-$ inclined on top of the $=$
answered Sep 8 '18 at 19:44
villasvvillasv
1487
$endgroup$
$begingroup$
The goal is to strictly make $0,0,0,0=8$ and not to "make the statement true" (therefore allowing this answer to be acceptable) :
$endgroup$
– user477343
Oct 3 '18 at 11:35
$begingroup$
When you say "the goal is to make [math expression]", then you're under the influence of the interpretation of that mathematical expression. IMO, "make the statement true" is the one and only interpretation to be taken, while you could argue more strict rules about not manipulating already existing symbols.
$endgroup$
– villasv
Oct 3 '18 at 13:18
$begingroup$
I see, you have a point. I should've said that my comment is solely based on how I interpreted the question, possibly explaining why your answer was sadly downvoted. I am positive that there exists another answer that manipulates the equation in a similar way that you have. Sorry if I sounded mean. You can get a $(+1)$ but in at least $2$ hours, once my daily voting limit (DVL) is over :)
$endgroup$
– user477343
Oct 3 '18 at 21:00
add a comment |
0
$begingroup$
Similar to @Vaelus
$0+0+0+0 leq 8$
Explanation
You can get the $leq$ by adding a $-$ inclined on top of the $=$
answered Sep 8 '18 at 19:44
villasvvillasv
1487
$endgroup$
$begingroup$
The goal is to strictly make $0,0,0,0=8$ and not to "make the statement true" (therefore allowing this answer to be acceptable) :
$endgroup$
– user477343
Oct 3 '18 at 11:35
$begingroup$
When you say "the goal is to make [math expression]", then you're under the influence of the interpretation of that mathematical expression. IMO, "make the statement true" is the one and only interpretation to be taken, while you could argue more strict rules about not manipulating already existing symbols.
$endgroup$
– villasv
Oct 3 '18 at 13:18
$begingroup$
I see, you have a point. I should've said that my comment is solely based on how I interpreted the question, possibly explaining why your answer was sadly downvoted. I am positive that there exists another answer that manipulates the equation in a similar way that you have. Sorry if I sounded mean. You can get a $(+1)$ but in at least $2$ hours, once my daily voting limit (DVL) is over :)
$endgroup$
– user477343
Oct 3 '18 at 21:00
add a comment |
0
0
0
$begingroup$
Similar to @Vaelus
$0+0+0+0 leq 8$
Explanation
You can get the $leq$ by adding a $-$ inclined on top of the $=$
answered Sep 8 '18 at 19:44
villasvvillasv
1487
$endgroup$
Similar to @Vaelus
$0+0+0+0 leq 8$
Explanation
You can get the $leq$ by adding a $-$ inclined on top of the $=$
answered Sep 8 '18 at 19:44
villasvvillasv
1487
answered Sep 8 '18 at 19:44
villasvvillasv
1487
answered Sep 8 '18 at 19:44
villasvvillasv
1487
answered Sep 8 '18 at 19:44
villasvvillasv
1487
1487
$begingroup$
The goal is to strictly make $0,0,0,0=8$ and not to "make the statement true" (therefore allowing this answer to be acceptable) :
$endgroup$
– user477343
Oct 3 '18 at 11:35
$begingroup$
When you say "the goal is to make [math expression]", then you're under the influence of the interpretation of that mathematical expression. IMO, "make the statement true" is the one and only interpretation to be taken, while you could argue more strict rules about not manipulating already existing symbols.
$endgroup$
– villasv
Oct 3 '18 at 13:18
$begingroup$
I see, you have a point. I should've said that my comment is solely based on how I interpreted the question, possibly explaining why your answer was sadly downvoted. I am positive that there exists another answer that manipulates the equation in a similar way that you have. Sorry if I sounded mean. You can get a $(+1)$ but in at least $2$ hours, once my daily voting limit (DVL) is over :)
$endgroup$
– user477343
Oct 3 '18 at 21:00
add a comment |
$begingroup$
The goal is to strictly make $0,0,0,0=8$ and not to "make the statement true" (therefore allowing this answer to be acceptable) :
$endgroup$
– user477343
Oct 3 '18 at 11:35
$begingroup$
When you say "the goal is to make [math expression]", then you're under the influence of the interpretation of that mathematical expression. IMO, "make the statement true" is the one and only interpretation to be taken, while you could argue more strict rules about not manipulating already existing symbols.
$endgroup$
– villasv
Oct 3 '18 at 13:18
$begingroup$
I see, you have a point. I should've said that my comment is solely based on how I interpreted the question, possibly explaining why your answer was sadly downvoted. I am positive that there exists another answer that manipulates the equation in a similar way that you have. Sorry if I sounded mean. You can get a $(+1)$ but in at least $2$ hours, once my daily voting limit (DVL) is over :)
$endgroup$
– user477343
Oct 3 '18 at 21:00
$begingroup$
The goal is to strictly make $0,0,0,0=8$ and not to "make the statement true" (therefore allowing this answer to be acceptable) :
$endgroup$
– user477343
Oct 3 '18 at 11:35
$begingroup$
The goal is to strictly make $0,0,0,0=8$ and not to "make the statement true" (therefore allowing this answer to be acceptable) :
$endgroup$
– user477343
Oct 3 '18 at 11:35
$begingroup$
When you say "the goal is to make [math expression]", then you're under the influence of the interpretation of that mathematical expression. IMO, "make the statement true" is the one and only interpretation to be taken, while you could argue more strict rules about not manipulating already existing symbols.
$endgroup$
– villasv
Oct 3 '18 at 13:18
$begingroup$
When you say "the goal is to make [math expression]", then you're under the influence of the interpretation of that mathematical expression. IMO, "make the statement true" is the one and only interpretation to be taken, while you could argue more strict rules about not manipulating already existing symbols.
$endgroup$
– villasv
Oct 3 '18 at 13:18
$begingroup$
I see, you have a point. I should've said that my comment is solely based on how I interpreted the question, possibly explaining why your answer was sadly downvoted. I am positive that there exists another answer that manipulates the equation in a similar way that you have. Sorry if I sounded mean. You can get a $(+1)$ but in at least $2$ hours, once my daily voting limit (DVL) is over :)
$endgroup$
– user477343
Oct 3 '18 at 21:00
$begingroup$
I see, you have a point. I should've said that my comment is solely based on how I interpreted the question, possibly explaining why your answer was sadly downvoted. I am positive that there exists another answer that manipulates the equation in a similar way that you have. Sorry if I sounded mean. You can get a $(+1)$ but in at least $2$ hours, once my daily voting limit (DVL) is over :)
$endgroup$
– user477343
Oct 3 '18 at 21:00
add a comment |
0
$begingroup$
As per the list of allowed symbols we are clearly allowed to use "$,$" and "$.$"
This is doubly evident as otherwise how would we use the $mathbb{concatenation}$ function without a comma to separate the arguments?
So the solution is:
$0! - mathbb{concatenation}(., 0 + 0! + 0!!) = .8$
$1 - mathbb{concatenation}(., 0 + 1 + 1) = .8$
$1 - mathbb{concatenation}(., 2) = .8$
$1 - .2 = .8$
answered Sep 10 '18 at 16:55
SamYonnouSamYonnou
69067
$endgroup$
$begingroup$
You may simplify0!!as0!.
$endgroup$
– Cœur
Sep 11 '18 at 15:28
$begingroup$
I felt like $0+0!+0!!$ looked more elegant/purposeful as if it was part of a sequence like $a(n, m) = a(n-1,m)!; a(0,m)=m$
$endgroup$
– SamYonnou
Sep 11 '18 at 16:58
add a comment |
0
$begingroup$
As per the list of allowed symbols we are clearly allowed to use "$,$" and "$.$"
This is doubly evident as otherwise how would we use the $mathbb{concatenation}$ function without a comma to separate the arguments?
So the solution is:
$0! - mathbb{concatenation}(., 0 + 0! + 0!!) = .8$
$1 - mathbb{concatenation}(., 0 + 1 + 1) = .8$
$1 - mathbb{concatenation}(., 2) = .8$
$1 - .2 = .8$
answered Sep 10 '18 at 16:55
SamYonnouSamYonnou
69067
$endgroup$
$begingroup$
You may simplify0!!as0!.
$endgroup$
– Cœur
Sep 11 '18 at 15:28
$begingroup$
I felt like $0+0!+0!!$ looked more elegant/purposeful as if it was part of a sequence like $a(n, m) = a(n-1,m)!; a(0,m)=m$
$endgroup$
– SamYonnou
Sep 11 '18 at 16:58
add a comment |
0
0
0
$begingroup$
As per the list of allowed symbols we are clearly allowed to use "$,$" and "$.$"
This is doubly evident as otherwise how would we use the $mathbb{concatenation}$ function without a comma to separate the arguments?
So the solution is:
$0! - mathbb{concatenation}(., 0 + 0! + 0!!) = .8$
$1 - mathbb{concatenation}(., 0 + 1 + 1) = .8$
$1 - mathbb{concatenation}(., 2) = .8$
$1 - .2 = .8$
answered Sep 10 '18 at 16:55
SamYonnouSamYonnou
69067
$endgroup$
As per the list of allowed symbols we are clearly allowed to use "$,$" and "$.$"
This is doubly evident as otherwise how would we use the $mathbb{concatenation}$ function without a comma to separate the arguments?
So the solution is:
$0! - mathbb{concatenation}(., 0 + 0! + 0!!) = .8$
$1 - mathbb{concatenation}(., 0 + 1 + 1) = .8$
$1 - mathbb{concatenation}(., 2) = .8$
$1 - .2 = .8$
answered Sep 10 '18 at 16:55
SamYonnouSamYonnou
69067
answered Sep 10 '18 at 16:55
SamYonnouSamYonnou
69067
answered Sep 10 '18 at 16:55
SamYonnouSamYonnou
69067
answered Sep 10 '18 at 16:55
SamYonnouSamYonnou
69067
69067
$begingroup$
You may simplify0!!as0!.
$endgroup$
– Cœur
Sep 11 '18 at 15:28
$begingroup$
I felt like $0+0!+0!!$ looked more elegant/purposeful as if it was part of a sequence like $a(n, m) = a(n-1,m)!; a(0,m)=m$
$endgroup$
– SamYonnou
Sep 11 '18 at 16:58
add a comment |
$begingroup$
You may simplify0!!as0!.
$endgroup$
– Cœur
Sep 11 '18 at 15:28
$begingroup$
I felt like $0+0!+0!!$ looked more elegant/purposeful as if it was part of a sequence like $a(n, m) = a(n-1,m)!; a(0,m)=m$
$endgroup$
– SamYonnou
Sep 11 '18 at 16:58
$begingroup$
You may simplify
0!! as 0!.$endgroup$
– Cœur
Sep 11 '18 at 15:28
$begingroup$
You may simplify
0!! as 0!.$endgroup$
– Cœur
Sep 11 '18 at 15:28
$begingroup$
I felt like $0+0!+0!!$ looked more elegant/purposeful as if it was part of a sequence like $a(n, m) = a(n-1,m)!; a(0,m)=m$
$endgroup$
– SamYonnou
Sep 11 '18 at 16:58
$begingroup$
I felt like $0+0!+0!!$ looked more elegant/purposeful as if it was part of a sequence like $a(n, m) = a(n-1,m)!; a(0,m)=m$
$endgroup$
– SamYonnou
Sep 11 '18 at 16:58
add a comment |
-1
$begingroup$
$00^{00} = 0^0$ which is indeterminate. In some sense, an indeterminate form can be equal to any value, since in Calculus, a function that approaches "$0^0$" can approach any real value, including $8$. So in that sense, $0^0 = 8$.
If you don't like concatenating two zeroes as "$00$", then $0^0 + 0 + 0$ also works.
answered Sep 7 '18 at 13:22
RothXRothX
1173
$endgroup$
7
$begingroup$
There are no widely accepted definitions under which your equation is considered to be true.
$endgroup$
– Tanner Swett
Sep 7 '18 at 17:34
$begingroup$
@TannerSwett Didn't think it was that unreasonable. My thinking is you can construct a function $f(x)^{g(x)}$ where $f$ and $g$ both limit to zero at some point $c$ such that $f(x)^{g(x)}$ limits to any real number at $c$, including $8$. Admittedly, writing $0^0 = 8$ is basically mathematically false. It is indeterminate for the reason I stated above. But this is Puzzling SE, not Mathematics SE, so I figured it was ok to think outside the box and not be formal with math.
$endgroup$
– RothX
Sep 10 '18 at 13:23
$begingroup$
Yeah, I dunno if I'd call it "that unreasonable". But if I saw a student say that the equation $0^0 = 8$ is true, I'd think that they probably misunderstand what an indeterminate form means. If I saw a math professor say the same thing, I'd ask them if they could please be a little more rigorous.
$endgroup$
– Tanner Swett
Sep 10 '18 at 13:59
$begingroup$
@TannerSwett Yeah, I completely agree. I'd never write $0^0 = 8$ in a mathematical setting, but since this is a puzzle, not a math problem, I thought maybe I could stretch a bit.
$endgroup$
– RothX
Sep 11 '18 at 1:58
add a comment |
-1
$begingroup$
$00^{00} = 0^0$ which is indeterminate. In some sense, an indeterminate form can be equal to any value, since in Calculus, a function that approaches "$0^0$" can approach any real value, including $8$. So in that sense, $0^0 = 8$.
If you don't like concatenating two zeroes as "$00$", then $0^0 + 0 + 0$ also works.
answered Sep 7 '18 at 13:22
RothXRothX
1173
$endgroup$
7
$begingroup$
There are no widely accepted definitions under which your equation is considered to be true.
$endgroup$
– Tanner Swett
Sep 7 '18 at 17:34
$begingroup$
@TannerSwett Didn't think it was that unreasonable. My thinking is you can construct a function $f(x)^{g(x)}$ where $f$ and $g$ both limit to zero at some point $c$ such that $f(x)^{g(x)}$ limits to any real number at $c$, including $8$. Admittedly, writing $0^0 = 8$ is basically mathematically false. It is indeterminate for the reason I stated above. But this is Puzzling SE, not Mathematics SE, so I figured it was ok to think outside the box and not be formal with math.
$endgroup$
– RothX
Sep 10 '18 at 13:23
$begingroup$
Yeah, I dunno if I'd call it "that unreasonable". But if I saw a student say that the equation $0^0 = 8$ is true, I'd think that they probably misunderstand what an indeterminate form means. If I saw a math professor say the same thing, I'd ask them if they could please be a little more rigorous.
$endgroup$
– Tanner Swett
Sep 10 '18 at 13:59
$begingroup$
@TannerSwett Yeah, I completely agree. I'd never write $0^0 = 8$ in a mathematical setting, but since this is a puzzle, not a math problem, I thought maybe I could stretch a bit.
$endgroup$
– RothX
Sep 11 '18 at 1:58
add a comment |
-1
-1
-1
$begingroup$
$00^{00} = 0^0$ which is indeterminate. In some sense, an indeterminate form can be equal to any value, since in Calculus, a function that approaches "$0^0$" can approach any real value, including $8$. So in that sense, $0^0 = 8$.
If you don't like concatenating two zeroes as "$00$", then $0^0 + 0 + 0$ also works.
answered Sep 7 '18 at 13:22
RothXRothX
1173
$endgroup$
$00^{00} = 0^0$ which is indeterminate. In some sense, an indeterminate form can be equal to any value, since in Calculus, a function that approaches "$0^0$" can approach any real value, including $8$. So in that sense, $0^0 = 8$.
If you don't like concatenating two zeroes as "$00$", then $0^0 + 0 + 0$ also works.
answered Sep 7 '18 at 13:22
RothXRothX
1173
edited Sep 7 '18 at 13:30
answered Sep 7 '18 at 13:22
RothXRothX
1173
answered Sep 7 '18 at 13:22
RothXRothX
1173
answered Sep 7 '18 at 13:22
RothXRothX
1173
1173
7
$begingroup$
There are no widely accepted definitions under which your equation is considered to be true.
$endgroup$
– Tanner Swett
Sep 7 '18 at 17:34
$begingroup$
@TannerSwett Didn't think it was that unreasonable. My thinking is you can construct a function $f(x)^{g(x)}$ where $f$ and $g$ both limit to zero at some point $c$ such that $f(x)^{g(x)}$ limits to any real number at $c$, including $8$. Admittedly, writing $0^0 = 8$ is basically mathematically false. It is indeterminate for the reason I stated above. But this is Puzzling SE, not Mathematics SE, so I figured it was ok to think outside the box and not be formal with math.
$endgroup$
– RothX
Sep 10 '18 at 13:23
$begingroup$
Yeah, I dunno if I'd call it "that unreasonable". But if I saw a student say that the equation $0^0 = 8$ is true, I'd think that they probably misunderstand what an indeterminate form means. If I saw a math professor say the same thing, I'd ask them if they could please be a little more rigorous.
$endgroup$
– Tanner Swett
Sep 10 '18 at 13:59
$begingroup$
@TannerSwett Yeah, I completely agree. I'd never write $0^0 = 8$ in a mathematical setting, but since this is a puzzle, not a math problem, I thought maybe I could stretch a bit.
$endgroup$
– RothX
Sep 11 '18 at 1:58
add a comment |
7
$begingroup$
There are no widely accepted definitions under which your equation is considered to be true.
$endgroup$
– Tanner Swett
Sep 7 '18 at 17:34
$begingroup$
@TannerSwett Didn't think it was that unreasonable. My thinking is you can construct a function $f(x)^{g(x)}$ where $f$ and $g$ both limit to zero at some point $c$ such that $f(x)^{g(x)}$ limits to any real number at $c$, including $8$. Admittedly, writing $0^0 = 8$ is basically mathematically false. It is indeterminate for the reason I stated above. But this is Puzzling SE, not Mathematics SE, so I figured it was ok to think outside the box and not be formal with math.
$endgroup$
– RothX
Sep 10 '18 at 13:23
$begingroup$
Yeah, I dunno if I'd call it "that unreasonable". But if I saw a student say that the equation $0^0 = 8$ is true, I'd think that they probably misunderstand what an indeterminate form means. If I saw a math professor say the same thing, I'd ask them if they could please be a little more rigorous.
$endgroup$
– Tanner Swett
Sep 10 '18 at 13:59
$begingroup$
@TannerSwett Yeah, I completely agree. I'd never write $0^0 = 8$ in a mathematical setting, but since this is a puzzle, not a math problem, I thought maybe I could stretch a bit.
$endgroup$
– RothX
Sep 11 '18 at 1:58
7
7
$begingroup$
There are no widely accepted definitions under which your equation is considered to be true.
$endgroup$
– Tanner Swett
Sep 7 '18 at 17:34
$begingroup$
There are no widely accepted definitions under which your equation is considered to be true.
$endgroup$
– Tanner Swett
Sep 7 '18 at 17:34
$begingroup$
@TannerSwett Didn't think it was that unreasonable. My thinking is you can construct a function $f(x)^{g(x)}$ where $f$ and $g$ both limit to zero at some point $c$ such that $f(x)^{g(x)}$ limits to any real number at $c$, including $8$. Admittedly, writing $0^0 = 8$ is basically mathematically false. It is indeterminate for the reason I stated above. But this is Puzzling SE, not Mathematics SE, so I figured it was ok to think outside the box and not be formal with math.
$endgroup$
– RothX
Sep 10 '18 at 13:23
$begingroup$
@TannerSwett Didn't think it was that unreasonable. My thinking is you can construct a function $f(x)^{g(x)}$ where $f$ and $g$ both limit to zero at some point $c$ such that $f(x)^{g(x)}$ limits to any real number at $c$, including $8$. Admittedly, writing $0^0 = 8$ is basically mathematically false. It is indeterminate for the reason I stated above. But this is Puzzling SE, not Mathematics SE, so I figured it was ok to think outside the box and not be formal with math.
$endgroup$
– RothX
Sep 10 '18 at 13:23
$begingroup$
Yeah, I dunno if I'd call it "that unreasonable". But if I saw a student say that the equation $0^0 = 8$ is true, I'd think that they probably misunderstand what an indeterminate form means. If I saw a math professor say the same thing, I'd ask them if they could please be a little more rigorous.
$endgroup$
– Tanner Swett
Sep 10 '18 at 13:59
$begingroup$
Yeah, I dunno if I'd call it "that unreasonable". But if I saw a student say that the equation $0^0 = 8$ is true, I'd think that they probably misunderstand what an indeterminate form means. If I saw a math professor say the same thing, I'd ask them if they could please be a little more rigorous.
$endgroup$
– Tanner Swett
Sep 10 '18 at 13:59
$begingroup$
@TannerSwett Yeah, I completely agree. I'd never write $0^0 = 8$ in a mathematical setting, but since this is a puzzle, not a math problem, I thought maybe I could stretch a bit.
$endgroup$
– RothX
Sep 11 '18 at 1:58
$begingroup$
@TannerSwett Yeah, I completely agree. I'd never write $0^0 = 8$ in a mathematical setting, but since this is a puzzle, not a math problem, I thought maybe I could stretch a bit.
$endgroup$
– RothX
Sep 11 '18 at 1:58
add a comment |
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40
$begingroup$
Just put a slash over the equal sign!
$endgroup$
– Yout Ried
Sep 7 '18 at 0:40
13
$begingroup$
@YoutRied: standard loophole: neither funny nor creative. It's usually the first answer on any math puzzle.
$endgroup$
– Thomas Weller
Sep 7 '18 at 18:40
2
$begingroup$
Are we allowed to move anything, or is it insert only?
$endgroup$
– ctrl-alt-delor
Sep 7 '18 at 19:33
2
$begingroup$
Are you allowed to concatenate operators? (e.g
!!,**,--, and++)$endgroup$
– Ole Tange
Sep 9 '18 at 1:09
1
$begingroup$
@OleTange The accepted answer does concatenate operators. Also, the question allows it.
$endgroup$
– haykam
Sep 9 '18 at 13:59