1 2 3 4 5 6 7 8 9 = 100
$begingroup$
The sequence of numbers $1 2 3 4 5 6 7 8 9$ has the property that you can insert mathematical operators in between the numbers from $1$ to $9$ and make the expression evaluate to 100. For example:
$$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 times 9 = 100$$
There are possibly hundreds of solutions to this problem, involving different varieties of operators. What is the expression with the fewest number of operators inserted (out of the set $+, -, times, div$ and maybe $sqrt{}$ and $!$) that evaluates to 100?
mathematics calculation-puzzle formation-of-numbers
edited May 13 '15 at 20:01
Rand al'Thor
71.1k14236472
asked Apr 13 '15 at 19:27
Joe Z.Joe Z.
21.5k675151
$endgroup$
This question has an open bounty worth +50
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 @IanMacDonald's answer; it is pretty creative, but unfortunately since it was made before a vital edit that changed the rules a bit, it was not the correct answer. Hence, I think it deserves a +50 rep bounty (as opposed to the usual +100). Nevertheless, it is still a pretty good answer! Well done! :P
|
show 14 more comments
$begingroup$
The sequence of numbers $1 2 3 4 5 6 7 8 9$ has the property that you can insert mathematical operators in between the numbers from $1$ to $9$ and make the expression evaluate to 100. For example:
$$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 times 9 = 100$$
There are possibly hundreds of solutions to this problem, involving different varieties of operators. What is the expression with the fewest number of operators inserted (out of the set $+, -, times, div$ and maybe $sqrt{}$ and $!$) that evaluates to 100?
mathematics calculation-puzzle formation-of-numbers
edited May 13 '15 at 20:01
Rand al'Thor
71.1k14236472
asked Apr 13 '15 at 19:27
Joe Z.Joe Z.
21.5k675151
$endgroup$
This question has an open bounty worth +50
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 @IanMacDonald's answer; it is pretty creative, but unfortunately since it was made before a vital edit that changed the rules a bit, it was not the correct answer. Hence, I think it deserves a +50 rep bounty (as opposed to the usual +100). Nevertheless, it is still a pretty good answer! Well done! :P
4
$begingroup$
Can I insert a symbol between = and 100?
$endgroup$
– Ben Frankel
Apr 13 '15 at 19:45
10
$begingroup$
@JoeZ. You should define the question such that you don't need to make ex post facto judgements on what is or isn't allowed. You failed to specify specifically which "mathematical symbols" are allowed and their meaning. Numbers are symbols, and they are used in mathematics. Taking it a step farther, I could invent a new operator with its own symbol (as is often done in mathematics) and define it to always evaluate to 100. Stick it anywhere on the left and voila: 100 = 100, one added symbol.
$endgroup$
– bcrist
Apr 14 '15 at 1:10
3
$begingroup$
@JoeZ. I wasn't complaining or implying that my hypothetical solution should be considered a valid solution to the puzzle, but trying to point out that if you explicitly listed the specific operators and their acceptable meanings (some operators have different or multiple meanings in different areas of mathematics) you would not need to contend with a plethora of "clever" solutions utilizing lateral thinking.
$endgroup$
– bcrist
Apr 14 '15 at 1:47
3
$begingroup$
@JoeZ. I believe my first comment was and still is quite constructive: it gives specific advice for how to describe the puzzle less ambiguously. Namely, by listing each symbol/operator which is acceptable and what meaning it has. What irritates me is when people misinterpret constructive criticism and cry foul instead.
$endgroup$
– bcrist
Apr 14 '15 at 2:07
2
$begingroup$
@bcrist I apologize. I got antsy about this because I've been accused of moving the goalposts in much the same way at PCG.SE (the code golf site) more than once, and it pattern-matches to people complaining about "chameleon questions" where people do indeed believe their loophole-abusing solutions should be considered valid even if they ruin the challenge, almost as a penalty to the question-asker for not making the question clear enough in the first place. Looking back at my first reply to you, some of my negativity was pretty uncalled for, so I've just deleted my comments.
$endgroup$
– Joe Z.
Apr 14 '15 at 5:50
|
show 14 more comments
$begingroup$
The sequence of numbers $1 2 3 4 5 6 7 8 9$ has the property that you can insert mathematical operators in between the numbers from $1$ to $9$ and make the expression evaluate to 100. For example:
$$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 times 9 = 100$$
There are possibly hundreds of solutions to this problem, involving different varieties of operators. What is the expression with the fewest number of operators inserted (out of the set $+, -, times, div$ and maybe $sqrt{}$ and $!$) that evaluates to 100?
mathematics calculation-puzzle formation-of-numbers
edited May 13 '15 at 20:01
Rand al'Thor
71.1k14236472
asked Apr 13 '15 at 19:27
Joe Z.Joe Z.
21.5k675151
$endgroup$
The sequence of numbers $1 2 3 4 5 6 7 8 9$ has the property that you can insert mathematical operators in between the numbers from $1$ to $9$ and make the expression evaluate to 100. For example:
$$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 times 9 = 100$$
There are possibly hundreds of solutions to this problem, involving different varieties of operators. What is the expression with the fewest number of operators inserted (out of the set $+, -, times, div$ and maybe $sqrt{}$ and $!$) that evaluates to 100?
mathematics calculation-puzzle formation-of-numbers
mathematics calculation-puzzle formation-of-numbers
edited May 13 '15 at 20:01
Rand al'Thor
71.1k14236472
asked Apr 13 '15 at 19:27
Joe Z.Joe Z.
21.5k675151
edited May 13 '15 at 20:01
Rand al'Thor
71.1k14236472
asked Apr 13 '15 at 19:27
Joe Z.Joe Z.
21.5k675151
edited May 13 '15 at 20:01
Rand al'Thor
71.1k14236472
edited May 13 '15 at 20:01
Rand al'Thor
71.1k14236472
edited May 13 '15 at 20:01
Rand al'Thor
71.1k14236472
71.1k14236472
asked Apr 13 '15 at 19:27
Joe Z.Joe Z.
21.5k675151
asked Apr 13 '15 at 19:27
Joe Z.Joe Z.
21.5k675151
asked Apr 13 '15 at 19:27
Joe Z.Joe Z.
21.5k675151
21.5k675151
This question has an open bounty worth +50
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 @IanMacDonald's answer; it is pretty creative, but unfortunately since it was made before a vital edit that changed the rules a bit, it was not the correct answer. Hence, I think it deserves a +50 rep bounty (as opposed to the usual +100). Nevertheless, it is still a pretty good answer! Well done! :P
This question has an open bounty worth +50
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 @IanMacDonald's answer; it is pretty creative, but unfortunately since it was made before a vital edit that changed the rules a bit, it was not the correct answer. Hence, I think it deserves a +50 rep bounty (as opposed to the usual +100). Nevertheless, it is still a pretty good answer! Well done! :P
4
$begingroup$
Can I insert a symbol between = and 100?
$endgroup$
– Ben Frankel
Apr 13 '15 at 19:45
10
$begingroup$
@JoeZ. You should define the question such that you don't need to make ex post facto judgements on what is or isn't allowed. You failed to specify specifically which "mathematical symbols" are allowed and their meaning. Numbers are symbols, and they are used in mathematics. Taking it a step farther, I could invent a new operator with its own symbol (as is often done in mathematics) and define it to always evaluate to 100. Stick it anywhere on the left and voila: 100 = 100, one added symbol.
$endgroup$
– bcrist
Apr 14 '15 at 1:10
3
$begingroup$
@JoeZ. I wasn't complaining or implying that my hypothetical solution should be considered a valid solution to the puzzle, but trying to point out that if you explicitly listed the specific operators and their acceptable meanings (some operators have different or multiple meanings in different areas of mathematics) you would not need to contend with a plethora of "clever" solutions utilizing lateral thinking.
$endgroup$
– bcrist
Apr 14 '15 at 1:47
3
$begingroup$
@JoeZ. I believe my first comment was and still is quite constructive: it gives specific advice for how to describe the puzzle less ambiguously. Namely, by listing each symbol/operator which is acceptable and what meaning it has. What irritates me is when people misinterpret constructive criticism and cry foul instead.
$endgroup$
– bcrist
Apr 14 '15 at 2:07
2
$begingroup$
@bcrist I apologize. I got antsy about this because I've been accused of moving the goalposts in much the same way at PCG.SE (the code golf site) more than once, and it pattern-matches to people complaining about "chameleon questions" where people do indeed believe their loophole-abusing solutions should be considered valid even if they ruin the challenge, almost as a penalty to the question-asker for not making the question clear enough in the first place. Looking back at my first reply to you, some of my negativity was pretty uncalled for, so I've just deleted my comments.
$endgroup$
– Joe Z.
Apr 14 '15 at 5:50
|
show 14 more comments
4
$begingroup$
Can I insert a symbol between = and 100?
$endgroup$
– Ben Frankel
Apr 13 '15 at 19:45
10
$begingroup$
@JoeZ. You should define the question such that you don't need to make ex post facto judgements on what is or isn't allowed. You failed to specify specifically which "mathematical symbols" are allowed and their meaning. Numbers are symbols, and they are used in mathematics. Taking it a step farther, I could invent a new operator with its own symbol (as is often done in mathematics) and define it to always evaluate to 100. Stick it anywhere on the left and voila: 100 = 100, one added symbol.
$endgroup$
– bcrist
Apr 14 '15 at 1:10
3
$begingroup$
@JoeZ. I wasn't complaining or implying that my hypothetical solution should be considered a valid solution to the puzzle, but trying to point out that if you explicitly listed the specific operators and their acceptable meanings (some operators have different or multiple meanings in different areas of mathematics) you would not need to contend with a plethora of "clever" solutions utilizing lateral thinking.
$endgroup$
– bcrist
Apr 14 '15 at 1:47
3
$begingroup$
@JoeZ. I believe my first comment was and still is quite constructive: it gives specific advice for how to describe the puzzle less ambiguously. Namely, by listing each symbol/operator which is acceptable and what meaning it has. What irritates me is when people misinterpret constructive criticism and cry foul instead.
$endgroup$
– bcrist
Apr 14 '15 at 2:07
2
$begingroup$
@bcrist I apologize. I got antsy about this because I've been accused of moving the goalposts in much the same way at PCG.SE (the code golf site) more than once, and it pattern-matches to people complaining about "chameleon questions" where people do indeed believe their loophole-abusing solutions should be considered valid even if they ruin the challenge, almost as a penalty to the question-asker for not making the question clear enough in the first place. Looking back at my first reply to you, some of my negativity was pretty uncalled for, so I've just deleted my comments.
$endgroup$
– Joe Z.
Apr 14 '15 at 5:50
4
4
$begingroup$
Can I insert a symbol between = and 100?
$endgroup$
– Ben Frankel
Apr 13 '15 at 19:45
$begingroup$
Can I insert a symbol between = and 100?
$endgroup$
– Ben Frankel
Apr 13 '15 at 19:45
10
10
$begingroup$
@JoeZ. You should define the question such that you don't need to make ex post facto judgements on what is or isn't allowed. You failed to specify specifically which "mathematical symbols" are allowed and their meaning. Numbers are symbols, and they are used in mathematics. Taking it a step farther, I could invent a new operator with its own symbol (as is often done in mathematics) and define it to always evaluate to 100. Stick it anywhere on the left and voila: 100 = 100, one added symbol.
$endgroup$
– bcrist
Apr 14 '15 at 1:10
$begingroup$
@JoeZ. You should define the question such that you don't need to make ex post facto judgements on what is or isn't allowed. You failed to specify specifically which "mathematical symbols" are allowed and their meaning. Numbers are symbols, and they are used in mathematics. Taking it a step farther, I could invent a new operator with its own symbol (as is often done in mathematics) and define it to always evaluate to 100. Stick it anywhere on the left and voila: 100 = 100, one added symbol.
$endgroup$
– bcrist
Apr 14 '15 at 1:10
3
3
$begingroup$
@JoeZ. I wasn't complaining or implying that my hypothetical solution should be considered a valid solution to the puzzle, but trying to point out that if you explicitly listed the specific operators and their acceptable meanings (some operators have different or multiple meanings in different areas of mathematics) you would not need to contend with a plethora of "clever" solutions utilizing lateral thinking.
$endgroup$
– bcrist
Apr 14 '15 at 1:47
$begingroup$
@JoeZ. I wasn't complaining or implying that my hypothetical solution should be considered a valid solution to the puzzle, but trying to point out that if you explicitly listed the specific operators and their acceptable meanings (some operators have different or multiple meanings in different areas of mathematics) you would not need to contend with a plethora of "clever" solutions utilizing lateral thinking.
$endgroup$
– bcrist
Apr 14 '15 at 1:47
3
3
$begingroup$
@JoeZ. I believe my first comment was and still is quite constructive: it gives specific advice for how to describe the puzzle less ambiguously. Namely, by listing each symbol/operator which is acceptable and what meaning it has. What irritates me is when people misinterpret constructive criticism and cry foul instead.
$endgroup$
– bcrist
Apr 14 '15 at 2:07
$begingroup$
@JoeZ. I believe my first comment was and still is quite constructive: it gives specific advice for how to describe the puzzle less ambiguously. Namely, by listing each symbol/operator which is acceptable and what meaning it has. What irritates me is when people misinterpret constructive criticism and cry foul instead.
$endgroup$
– bcrist
Apr 14 '15 at 2:07
2
2
$begingroup$
@bcrist I apologize. I got antsy about this because I've been accused of moving the goalposts in much the same way at PCG.SE (the code golf site) more than once, and it pattern-matches to people complaining about "chameleon questions" where people do indeed believe their loophole-abusing solutions should be considered valid even if they ruin the challenge, almost as a penalty to the question-asker for not making the question clear enough in the first place. Looking back at my first reply to you, some of my negativity was pretty uncalled for, so I've just deleted my comments.
$endgroup$
– Joe Z.
Apr 14 '15 at 5:50
$begingroup$
@bcrist I apologize. I got antsy about this because I've been accused of moving the goalposts in much the same way at PCG.SE (the code golf site) more than once, and it pattern-matches to people complaining about "chameleon questions" where people do indeed believe their loophole-abusing solutions should be considered valid even if they ruin the challenge, almost as a penalty to the question-asker for not making the question clear enough in the first place. Looking back at my first reply to you, some of my negativity was pretty uncalled for, so I've just deleted my comments.
$endgroup$
– Joe Z.
Apr 14 '15 at 5:50
|
show 14 more comments
7 Answers
7
active
oldest
votes
$begingroup$
I believe that this is the smallest:
$123-45-67+89 = 100$
edited May 1 '18 at 4:25
kraby15
2,4143930
answered Apr 13 '15 at 19:36
Andrew SmithAndrew Smith
1,59489
$endgroup$
$begingroup$
This is the smallest one I know of too, but let's see if someone can beat it.
$endgroup$
– Joe Z.
Apr 13 '15 at 19:37
1
$begingroup$
This is the best, according to worldofnumbers.com/ninedig1.htm
$endgroup$
– JLee
Apr 13 '15 at 20:13
27
$begingroup$
Unless something like 123456789>=100 counts, I couldn't find a better one either
$endgroup$
– Andrew Smith
Apr 13 '15 at 20:14
1
$begingroup$
Verified shortest possible solution.
$endgroup$
– Oliver Ni
Dec 18 '16 at 18:20
add a comment |
$begingroup$
Note: This answer only applies prior to the edit that clarifies that the expression on the left must evaluate to 100, rather than simply the equation being true.
If you allow exponents, you can get away with just two:
$ 1^{ 23456789 } = 10^0 $
edited Apr 14 '15 at 17:33
Joe Z.
21.5k675151
answered Apr 13 '15 at 21:39
Ian MacDonaldIan MacDonald
11.3k2662
$endgroup$
2
$begingroup$
Well, you're not supposed to be able to insert anything between the digits of 100, but +1 for creativity.
$endgroup$
– Joe Z.
Apr 14 '15 at 3:17
1
$begingroup$
@EngineerToast 00 is not nonsense, though. Surely you understand it (and therefore it makes sense).
$endgroup$
– Ian MacDonald
Apr 14 '15 at 12:17
8
$begingroup$
@EngineerToast It would be 2 operations since you could write1^23456789=10^0
$endgroup$
– pacoverflow
Apr 14 '15 at 16:01
1
$begingroup$
Also, $0^0$ is undefined.
$endgroup$
– Joe Z.
Apr 14 '15 at 16:28
1
$begingroup$
@pacoverflow +1 Good optimization.
$endgroup$
– Engineer Toast
Apr 14 '15 at 22:06
|
show 5 more comments
$begingroup$
Best I can do so far is tie the smallest (3 operators):
1234%567%8^9 = 100
which Wolfram Alpha interprets as:
$(1234bmod567)bmod8^9 = 100$
$(100)bmod8^9 = 100$
For those who are unfamiliar, % is the modulo operator. Throw the equation in wolfram alpha.
EDIT:
Started on finding this solution before the operator limit was imposed. Not valid under the rules of the current question.
answered Apr 14 '15 at 18:21
Mwr247Mwr247
47535
$endgroup$
$begingroup$
How exactly does that work? I can't figure out which % applies first.
$endgroup$
– Joe Z.
Apr 14 '15 at 18:22
4
$begingroup$
Modulus works at the same level as multiplication/division in the order of operations. As such, no parenthesis are needed. The exponent is resolved first, then we work left to right with the modulus signs.
$endgroup$
– Mwr247
Apr 14 '15 at 18:27
1
$begingroup$
@Mwr247: I know what you mean. My answer was posted before the edit, too. People still keep upvoting it, though. -_-
$endgroup$
– Ian MacDonald
Apr 14 '15 at 18:36
1
$begingroup$
@IanMacDonald To be fair, yours was particularly clever hehe... Anyways, unless the rules change to allow modulo (which it totally should, given how ubiquitous it is), I've added a note to my answer indicating that it's not technically valid. Thanks again everyone =)
$endgroup$
– Mwr247
Apr 14 '15 at 18:40
1
$begingroup$
@jjdb Parentheses are used to override the normal order of operations, and themselves have a place in that order (albeit at the top). When not specified, the implicit order is to solve from left to right. Parentheses are therefore not required, but rather, recommended for clarity. The case you gave as an example, without parenthesis, would evaluate as: 78 / 10 / 7 = 1.1143... approx.
$endgroup$
– Mwr247
Apr 16 '15 at 13:37
|
show 6 more comments
$begingroup$
Addendum too long for a comment
I can confirm there is no better solution than Andrew Smith's if we use the operators +-*/.^ There is only one solution using 3 operators. There are two using 4:
123 +45 - 67 + 8 - 9
123 + 4- 5 + 67 - 89
There are 215 possibilities total and the number of solutions for the number of operators breaks down as follows:
Operators Solutions
0 0
1 0
2 0
3 1
4 2
5 9
6 57
7 104
8 42
We could throw in some more operators though I doubt they would help. I didn't try parentheses, floors, mod, sqrt, etc. If anyone wants a go, though, here's some (unoptimized) VBA that will brute force the basic operators in Excel:
Sub doTheMath()
Const op As String = " +-*/.^"
Dim s As String
Dim a, b, c, d, e, f, g, h
Dim i, j
Dim v
'Loop all possibilities
i = 2
j = 1
For a = 1 To Len(op): For b = 1 To Len(op): For c = 1 To Len(op): For d = 1 To Len(op)
For e = 1 To Len(op): For f = 1 To Len(op): For g = 1 To Len(op): For h = 1 To Len(op)
s = 1 & Mid(op, a, 1) & 2 & Mid(op, b, 1) & 3 & Mid(op, c, 1) & 4 & Mid(op, d, 1) & 5 & _
Mid(op, e, 1) & 6 & Mid(op, f, 1) & 7 & Mid(op, g, 1) & 8 & Mid(op, h, 1) & 9
v = Application.Evaluate(Replace(s, " ", ""))
If CStr(v) = CStr(100) Then
Cells(i, 1).Value = s
Cells(i, 2).Value = v
i = i + 1
End If
j = j + 1
If j > 1000000 Then
ThisWorkbook.Save
j = 1
End If
Next: Next: Next: Next: Next: Next: Next: Next
End Sub
answered Apr 13 '15 at 20:34
Engineer ToastEngineer Toast
11.7k136131
$endgroup$
add a comment |
$begingroup$
This is my 2nd attempt.
This time I'm using 7 operations.
I couldn't find any shorter than this, for now.
-(12+34-56)*(-7+8+9)
I've tried and tried and tried.
This simply multiplies 10 by 10.
Nothing fancy, sadly too long, nothing that can't be done easily...
But well, it solves the case!
answered Apr 14 '15 at 19:10
Ismael MiguelIsmael Miguel
302212
$endgroup$
add a comment |
$begingroup$
I know that technically I do not have a shorter solution, but I applied the modulo operator and found an answer!
123456789 % 123456689 = 100
and so on. The program I used:
#include <iostream>
int
main()
{
for (int i = 100; i < 999999999; i++)
{
if (123456789 % i == 100)
{
std::cout << i << std::endl;
}
}
}
answered Apr 14 '15 at 11:22
TheolodisTheolodis
1254
$endgroup$
6
$begingroup$
Welcome to Puzzling SE! Your answer seems to be a bit off-topic, you aren't supposed to use additional numbers than those given by the author!
$endgroup$
– leoll2
Apr 14 '15 at 11:38
$begingroup$
@leoll2 oops, I did read "what is the fewest number or operators"
$endgroup$
– Theolodis
Apr 14 '15 at 14:06
$begingroup$
of, not or. Don't worry, it happens :)
$endgroup$
– Joe Z.
Apr 14 '15 at 16:11
$begingroup$
123456789 % 169 = 92, not 100.
$endgroup$
– Engineer Toast
Apr 18 '18 at 13:48
add a comment |
$begingroup$
I'm not sure this is what you're looking for, but 123456789=/=100
This answer adds two symbols, "=" and "/".
answered Apr 14 '15 at 0:18
DiscOHDiscOH
44129
$endgroup$
2
$begingroup$
You can't do that, though. The symbol=/=doesn't mean anything, and even if you were trying to change it to $neq$, that's specifically disallowed.
$endgroup$
– Joe Z.
Apr 14 '15 at 0:20
$begingroup$
Would you be happier with "123456789 != 100"? ! meaning "not" is listed in wikipedia's big list of math symbols en.wikipedia.org/wiki/List_of_mathematical_symbols
$endgroup$
– DiscOH
Apr 14 '15 at 1:03
10
$begingroup$
Well, the problem is with the word "equation". Sure you've made the statement true, but it's no longer an equation.
$endgroup$
– Joe Z.
Apr 14 '15 at 1:37
$begingroup$
@JoeZ. An equation is an equality with at least one variable. What you mean is an equality.
$endgroup$
– user26486
Apr 14 '15 at 23:23
$begingroup$
Well, in any case I've edited the question statement again and that removes any ambiguity with regards to my choice of word when it comes to "equation".
$endgroup$
– Joe Z.
Apr 15 '15 at 0:04
add a comment |
protected by Community♦ Apr 14 '15 at 19:00
Thank you for your interest in this question.
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7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I believe that this is the smallest:
$123-45-67+89 = 100$
edited May 1 '18 at 4:25
kraby15
2,4143930
answered Apr 13 '15 at 19:36
Andrew SmithAndrew Smith
1,59489
$endgroup$
$begingroup$
This is the smallest one I know of too, but let's see if someone can beat it.
$endgroup$
– Joe Z.
Apr 13 '15 at 19:37
1
$begingroup$
This is the best, according to worldofnumbers.com/ninedig1.htm
$endgroup$
– JLee
Apr 13 '15 at 20:13
27
$begingroup$
Unless something like 123456789>=100 counts, I couldn't find a better one either
$endgroup$
– Andrew Smith
Apr 13 '15 at 20:14
1
$begingroup$
Verified shortest possible solution.
$endgroup$
– Oliver Ni
Dec 18 '16 at 18:20
add a comment |
$begingroup$
I believe that this is the smallest:
$123-45-67+89 = 100$
edited May 1 '18 at 4:25
kraby15
2,4143930
answered Apr 13 '15 at 19:36
Andrew SmithAndrew Smith
1,59489
$endgroup$
$begingroup$
This is the smallest one I know of too, but let's see if someone can beat it.
$endgroup$
– Joe Z.
Apr 13 '15 at 19:37
1
$begingroup$
This is the best, according to worldofnumbers.com/ninedig1.htm
$endgroup$
– JLee
Apr 13 '15 at 20:13
27
$begingroup$
Unless something like 123456789>=100 counts, I couldn't find a better one either
$endgroup$
– Andrew Smith
Apr 13 '15 at 20:14
1
$begingroup$
Verified shortest possible solution.
$endgroup$
– Oliver Ni
Dec 18 '16 at 18:20
add a comment |
$begingroup$
I believe that this is the smallest:
$123-45-67+89 = 100$
edited May 1 '18 at 4:25
kraby15
2,4143930
answered Apr 13 '15 at 19:36
Andrew SmithAndrew Smith
1,59489
$endgroup$
I believe that this is the smallest:
$123-45-67+89 = 100$
edited May 1 '18 at 4:25
kraby15
2,4143930
answered Apr 13 '15 at 19:36
Andrew SmithAndrew Smith
1,59489
edited May 1 '18 at 4:25
kraby15
2,4143930
edited May 1 '18 at 4:25
kraby15
2,4143930
edited May 1 '18 at 4:25
kraby15
2,4143930
2,4143930
answered Apr 13 '15 at 19:36
Andrew SmithAndrew Smith
1,59489
answered Apr 13 '15 at 19:36
Andrew SmithAndrew Smith
1,59489
answered Apr 13 '15 at 19:36
Andrew SmithAndrew Smith
1,59489
1,59489
$begingroup$
This is the smallest one I know of too, but let's see if someone can beat it.
$endgroup$
– Joe Z.
Apr 13 '15 at 19:37
1
$begingroup$
This is the best, according to worldofnumbers.com/ninedig1.htm
$endgroup$
– JLee
Apr 13 '15 at 20:13
27
$begingroup$
Unless something like 123456789>=100 counts, I couldn't find a better one either
$endgroup$
– Andrew Smith
Apr 13 '15 at 20:14
1
$begingroup$
Verified shortest possible solution.
$endgroup$
– Oliver Ni
Dec 18 '16 at 18:20
add a comment |
$begingroup$
This is the smallest one I know of too, but let's see if someone can beat it.
$endgroup$
– Joe Z.
Apr 13 '15 at 19:37
1
$begingroup$
This is the best, according to worldofnumbers.com/ninedig1.htm
$endgroup$
– JLee
Apr 13 '15 at 20:13
27
$begingroup$
Unless something like 123456789>=100 counts, I couldn't find a better one either
$endgroup$
– Andrew Smith
Apr 13 '15 at 20:14
1
$begingroup$
Verified shortest possible solution.
$endgroup$
– Oliver Ni
Dec 18 '16 at 18:20
$begingroup$
This is the smallest one I know of too, but let's see if someone can beat it.
$endgroup$
– Joe Z.
Apr 13 '15 at 19:37
$begingroup$
This is the smallest one I know of too, but let's see if someone can beat it.
$endgroup$
– Joe Z.
Apr 13 '15 at 19:37
1
1
$begingroup$
This is the best, according to worldofnumbers.com/ninedig1.htm
$endgroup$
– JLee
Apr 13 '15 at 20:13
$begingroup$
This is the best, according to worldofnumbers.com/ninedig1.htm
$endgroup$
– JLee
Apr 13 '15 at 20:13
27
27
$begingroup$
Unless something like 123456789>=100 counts, I couldn't find a better one either
$endgroup$
– Andrew Smith
Apr 13 '15 at 20:14
$begingroup$
Unless something like 123456789>=100 counts, I couldn't find a better one either
$endgroup$
– Andrew Smith
Apr 13 '15 at 20:14
1
1
$begingroup$
Verified shortest possible solution.
$endgroup$
– Oliver Ni
Dec 18 '16 at 18:20
$begingroup$
Verified shortest possible solution.
$endgroup$
– Oliver Ni
Dec 18 '16 at 18:20
add a comment |
$begingroup$
Note: This answer only applies prior to the edit that clarifies that the expression on the left must evaluate to 100, rather than simply the equation being true.
If you allow exponents, you can get away with just two:
$ 1^{ 23456789 } = 10^0 $
edited Apr 14 '15 at 17:33
Joe Z.
21.5k675151
answered Apr 13 '15 at 21:39
Ian MacDonaldIan MacDonald
11.3k2662
$endgroup$
2
$begingroup$
Well, you're not supposed to be able to insert anything between the digits of 100, but +1 for creativity.
$endgroup$
– Joe Z.
Apr 14 '15 at 3:17
1
$begingroup$
@EngineerToast 00 is not nonsense, though. Surely you understand it (and therefore it makes sense).
$endgroup$
– Ian MacDonald
Apr 14 '15 at 12:17
8
$begingroup$
@EngineerToast It would be 2 operations since you could write1^23456789=10^0
$endgroup$
– pacoverflow
Apr 14 '15 at 16:01
1
$begingroup$
Also, $0^0$ is undefined.
$endgroup$
– Joe Z.
Apr 14 '15 at 16:28
1
$begingroup$
@pacoverflow +1 Good optimization.
$endgroup$
– Engineer Toast
Apr 14 '15 at 22:06
|
show 5 more comments
$begingroup$
Note: This answer only applies prior to the edit that clarifies that the expression on the left must evaluate to 100, rather than simply the equation being true.
If you allow exponents, you can get away with just two:
$ 1^{ 23456789 } = 10^0 $
edited Apr 14 '15 at 17:33
Joe Z.
21.5k675151
answered Apr 13 '15 at 21:39
Ian MacDonaldIan MacDonald
11.3k2662
$endgroup$
2
$begingroup$
Well, you're not supposed to be able to insert anything between the digits of 100, but +1 for creativity.
$endgroup$
– Joe Z.
Apr 14 '15 at 3:17
1
$begingroup$
@EngineerToast 00 is not nonsense, though. Surely you understand it (and therefore it makes sense).
$endgroup$
– Ian MacDonald
Apr 14 '15 at 12:17
8
$begingroup$
@EngineerToast It would be 2 operations since you could write1^23456789=10^0
$endgroup$
– pacoverflow
Apr 14 '15 at 16:01
1
$begingroup$
Also, $0^0$ is undefined.
$endgroup$
– Joe Z.
Apr 14 '15 at 16:28
1
$begingroup$
@pacoverflow +1 Good optimization.
$endgroup$
– Engineer Toast
Apr 14 '15 at 22:06
|
show 5 more comments
$begingroup$
Note: This answer only applies prior to the edit that clarifies that the expression on the left must evaluate to 100, rather than simply the equation being true.
If you allow exponents, you can get away with just two:
$ 1^{ 23456789 } = 10^0 $
edited Apr 14 '15 at 17:33
Joe Z.
21.5k675151
answered Apr 13 '15 at 21:39
Ian MacDonaldIan MacDonald
11.3k2662
$endgroup$
Note: This answer only applies prior to the edit that clarifies that the expression on the left must evaluate to 100, rather than simply the equation being true.
If you allow exponents, you can get away with just two:
$ 1^{ 23456789 } = 10^0 $
edited Apr 14 '15 at 17:33
Joe Z.
21.5k675151
answered Apr 13 '15 at 21:39
Ian MacDonaldIan MacDonald
11.3k2662
edited Apr 14 '15 at 17:33
Joe Z.
21.5k675151
edited Apr 14 '15 at 17:33
Joe Z.
21.5k675151
edited Apr 14 '15 at 17:33
Joe Z.
21.5k675151
21.5k675151
answered Apr 13 '15 at 21:39
Ian MacDonaldIan MacDonald
11.3k2662
answered Apr 13 '15 at 21:39
Ian MacDonaldIan MacDonald
11.3k2662
answered Apr 13 '15 at 21:39
Ian MacDonaldIan MacDonald
11.3k2662
11.3k2662
2
$begingroup$
Well, you're not supposed to be able to insert anything between the digits of 100, but +1 for creativity.
$endgroup$
– Joe Z.
Apr 14 '15 at 3:17
1
$begingroup$
@EngineerToast 00 is not nonsense, though. Surely you understand it (and therefore it makes sense).
$endgroup$
– Ian MacDonald
Apr 14 '15 at 12:17
8
$begingroup$
@EngineerToast It would be 2 operations since you could write1^23456789=10^0
$endgroup$
– pacoverflow
Apr 14 '15 at 16:01
1
$begingroup$
Also, $0^0$ is undefined.
$endgroup$
– Joe Z.
Apr 14 '15 at 16:28
1
$begingroup$
@pacoverflow +1 Good optimization.
$endgroup$
– Engineer Toast
Apr 14 '15 at 22:06
|
show 5 more comments
2
$begingroup$
Well, you're not supposed to be able to insert anything between the digits of 100, but +1 for creativity.
$endgroup$
– Joe Z.
Apr 14 '15 at 3:17
1
$begingroup$
@EngineerToast 00 is not nonsense, though. Surely you understand it (and therefore it makes sense).
$endgroup$
– Ian MacDonald
Apr 14 '15 at 12:17
8
$begingroup$
@EngineerToast It would be 2 operations since you could write1^23456789=10^0
$endgroup$
– pacoverflow
Apr 14 '15 at 16:01
1
$begingroup$
Also, $0^0$ is undefined.
$endgroup$
– Joe Z.
Apr 14 '15 at 16:28
1
$begingroup$
@pacoverflow +1 Good optimization.
$endgroup$
– Engineer Toast
Apr 14 '15 at 22:06
2
2
$begingroup$
Well, you're not supposed to be able to insert anything between the digits of 100, but +1 for creativity.
$endgroup$
– Joe Z.
Apr 14 '15 at 3:17
$begingroup$
Well, you're not supposed to be able to insert anything between the digits of 100, but +1 for creativity.
$endgroup$
– Joe Z.
Apr 14 '15 at 3:17
1
1
$begingroup$
@EngineerToast 00 is not nonsense, though. Surely you understand it (and therefore it makes sense).
$endgroup$
– Ian MacDonald
Apr 14 '15 at 12:17
$begingroup$
@EngineerToast 00 is not nonsense, though. Surely you understand it (and therefore it makes sense).
$endgroup$
– Ian MacDonald
Apr 14 '15 at 12:17
8
8
$begingroup$
@EngineerToast It would be 2 operations since you could write
1^23456789=10^0$endgroup$
– pacoverflow
Apr 14 '15 at 16:01
$begingroup$
@EngineerToast It would be 2 operations since you could write
1^23456789=10^0$endgroup$
– pacoverflow
Apr 14 '15 at 16:01
1
1
$begingroup$
Also, $0^0$ is undefined.
$endgroup$
– Joe Z.
Apr 14 '15 at 16:28
$begingroup$
Also, $0^0$ is undefined.
$endgroup$
– Joe Z.
Apr 14 '15 at 16:28
1
1
$begingroup$
@pacoverflow +1 Good optimization.
$endgroup$
– Engineer Toast
Apr 14 '15 at 22:06
$begingroup$
@pacoverflow +1 Good optimization.
$endgroup$
– Engineer Toast
Apr 14 '15 at 22:06
|
show 5 more comments
$begingroup$
Best I can do so far is tie the smallest (3 operators):
1234%567%8^9 = 100
which Wolfram Alpha interprets as:
$(1234bmod567)bmod8^9 = 100$
$(100)bmod8^9 = 100$
For those who are unfamiliar, % is the modulo operator. Throw the equation in wolfram alpha.
EDIT:
Started on finding this solution before the operator limit was imposed. Not valid under the rules of the current question.
answered Apr 14 '15 at 18:21
Mwr247Mwr247
47535
$endgroup$
$begingroup$
How exactly does that work? I can't figure out which % applies first.
$endgroup$
– Joe Z.
Apr 14 '15 at 18:22
4
$begingroup$
Modulus works at the same level as multiplication/division in the order of operations. As such, no parenthesis are needed. The exponent is resolved first, then we work left to right with the modulus signs.
$endgroup$
– Mwr247
Apr 14 '15 at 18:27
1
$begingroup$
@Mwr247: I know what you mean. My answer was posted before the edit, too. People still keep upvoting it, though. -_-
$endgroup$
– Ian MacDonald
Apr 14 '15 at 18:36
1
$begingroup$
@IanMacDonald To be fair, yours was particularly clever hehe... Anyways, unless the rules change to allow modulo (which it totally should, given how ubiquitous it is), I've added a note to my answer indicating that it's not technically valid. Thanks again everyone =)
$endgroup$
– Mwr247
Apr 14 '15 at 18:40
1
$begingroup$
@jjdb Parentheses are used to override the normal order of operations, and themselves have a place in that order (albeit at the top). When not specified, the implicit order is to solve from left to right. Parentheses are therefore not required, but rather, recommended for clarity. The case you gave as an example, without parenthesis, would evaluate as: 78 / 10 / 7 = 1.1143... approx.
$endgroup$
– Mwr247
Apr 16 '15 at 13:37
|
show 6 more comments
$begingroup$
Best I can do so far is tie the smallest (3 operators):
1234%567%8^9 = 100
which Wolfram Alpha interprets as:
$(1234bmod567)bmod8^9 = 100$
$(100)bmod8^9 = 100$
For those who are unfamiliar, % is the modulo operator. Throw the equation in wolfram alpha.
EDIT:
Started on finding this solution before the operator limit was imposed. Not valid under the rules of the current question.
answered Apr 14 '15 at 18:21
Mwr247Mwr247
47535
$endgroup$
$begingroup$
How exactly does that work? I can't figure out which % applies first.
$endgroup$
– Joe Z.
Apr 14 '15 at 18:22
4
$begingroup$
Modulus works at the same level as multiplication/division in the order of operations. As such, no parenthesis are needed. The exponent is resolved first, then we work left to right with the modulus signs.
$endgroup$
– Mwr247
Apr 14 '15 at 18:27
1
$begingroup$
@Mwr247: I know what you mean. My answer was posted before the edit, too. People still keep upvoting it, though. -_-
$endgroup$
– Ian MacDonald
Apr 14 '15 at 18:36
1
$begingroup$
@IanMacDonald To be fair, yours was particularly clever hehe... Anyways, unless the rules change to allow modulo (which it totally should, given how ubiquitous it is), I've added a note to my answer indicating that it's not technically valid. Thanks again everyone =)
$endgroup$
– Mwr247
Apr 14 '15 at 18:40
1
$begingroup$
@jjdb Parentheses are used to override the normal order of operations, and themselves have a place in that order (albeit at the top). When not specified, the implicit order is to solve from left to right. Parentheses are therefore not required, but rather, recommended for clarity. The case you gave as an example, without parenthesis, would evaluate as: 78 / 10 / 7 = 1.1143... approx.
$endgroup$
– Mwr247
Apr 16 '15 at 13:37
|
show 6 more comments
$begingroup$
Best I can do so far is tie the smallest (3 operators):
1234%567%8^9 = 100
which Wolfram Alpha interprets as:
$(1234bmod567)bmod8^9 = 100$
$(100)bmod8^9 = 100$
For those who are unfamiliar, % is the modulo operator. Throw the equation in wolfram alpha.
EDIT:
Started on finding this solution before the operator limit was imposed. Not valid under the rules of the current question.
answered Apr 14 '15 at 18:21
Mwr247Mwr247
47535
$endgroup$
Best I can do so far is tie the smallest (3 operators):
1234%567%8^9 = 100
which Wolfram Alpha interprets as:
$(1234bmod567)bmod8^9 = 100$
$(100)bmod8^9 = 100$
For those who are unfamiliar, % is the modulo operator. Throw the equation in wolfram alpha.
EDIT:
Started on finding this solution before the operator limit was imposed. Not valid under the rules of the current question.
answered Apr 14 '15 at 18:21
Mwr247Mwr247
47535
edited Apr 14 '15 at 18:37
answered Apr 14 '15 at 18:21
Mwr247Mwr247
47535
answered Apr 14 '15 at 18:21
Mwr247Mwr247
47535
answered Apr 14 '15 at 18:21
Mwr247Mwr247
47535
47535
$begingroup$
How exactly does that work? I can't figure out which % applies first.
$endgroup$
– Joe Z.
Apr 14 '15 at 18:22
4
$begingroup$
Modulus works at the same level as multiplication/division in the order of operations. As such, no parenthesis are needed. The exponent is resolved first, then we work left to right with the modulus signs.
$endgroup$
– Mwr247
Apr 14 '15 at 18:27
1
$begingroup$
@Mwr247: I know what you mean. My answer was posted before the edit, too. People still keep upvoting it, though. -_-
$endgroup$
– Ian MacDonald
Apr 14 '15 at 18:36
1
$begingroup$
@IanMacDonald To be fair, yours was particularly clever hehe... Anyways, unless the rules change to allow modulo (which it totally should, given how ubiquitous it is), I've added a note to my answer indicating that it's not technically valid. Thanks again everyone =)
$endgroup$
– Mwr247
Apr 14 '15 at 18:40
1
$begingroup$
@jjdb Parentheses are used to override the normal order of operations, and themselves have a place in that order (albeit at the top). When not specified, the implicit order is to solve from left to right. Parentheses are therefore not required, but rather, recommended for clarity. The case you gave as an example, without parenthesis, would evaluate as: 78 / 10 / 7 = 1.1143... approx.
$endgroup$
– Mwr247
Apr 16 '15 at 13:37
|
show 6 more comments
$begingroup$
How exactly does that work? I can't figure out which % applies first.
$endgroup$
– Joe Z.
Apr 14 '15 at 18:22
4
$begingroup$
Modulus works at the same level as multiplication/division in the order of operations. As such, no parenthesis are needed. The exponent is resolved first, then we work left to right with the modulus signs.
$endgroup$
– Mwr247
Apr 14 '15 at 18:27
1
$begingroup$
@Mwr247: I know what you mean. My answer was posted before the edit, too. People still keep upvoting it, though. -_-
$endgroup$
– Ian MacDonald
Apr 14 '15 at 18:36
1
$begingroup$
@IanMacDonald To be fair, yours was particularly clever hehe... Anyways, unless the rules change to allow modulo (which it totally should, given how ubiquitous it is), I've added a note to my answer indicating that it's not technically valid. Thanks again everyone =)
$endgroup$
– Mwr247
Apr 14 '15 at 18:40
1
$begingroup$
@jjdb Parentheses are used to override the normal order of operations, and themselves have a place in that order (albeit at the top). When not specified, the implicit order is to solve from left to right. Parentheses are therefore not required, but rather, recommended for clarity. The case you gave as an example, without parenthesis, would evaluate as: 78 / 10 / 7 = 1.1143... approx.
$endgroup$
– Mwr247
Apr 16 '15 at 13:37
$begingroup$
How exactly does that work? I can't figure out which % applies first.
$endgroup$
– Joe Z.
Apr 14 '15 at 18:22
$begingroup$
How exactly does that work? I can't figure out which % applies first.
$endgroup$
– Joe Z.
Apr 14 '15 at 18:22
4
4
$begingroup$
Modulus works at the same level as multiplication/division in the order of operations. As such, no parenthesis are needed. The exponent is resolved first, then we work left to right with the modulus signs.
$endgroup$
– Mwr247
Apr 14 '15 at 18:27
$begingroup$
Modulus works at the same level as multiplication/division in the order of operations. As such, no parenthesis are needed. The exponent is resolved first, then we work left to right with the modulus signs.
$endgroup$
– Mwr247
Apr 14 '15 at 18:27
1
1
$begingroup$
@Mwr247: I know what you mean. My answer was posted before the edit, too. People still keep upvoting it, though. -_-
$endgroup$
– Ian MacDonald
Apr 14 '15 at 18:36
$begingroup$
@Mwr247: I know what you mean. My answer was posted before the edit, too. People still keep upvoting it, though. -_-
$endgroup$
– Ian MacDonald
Apr 14 '15 at 18:36
1
1
$begingroup$
@IanMacDonald To be fair, yours was particularly clever hehe... Anyways, unless the rules change to allow modulo (which it totally should, given how ubiquitous it is), I've added a note to my answer indicating that it's not technically valid. Thanks again everyone =)
$endgroup$
– Mwr247
Apr 14 '15 at 18:40
$begingroup$
@IanMacDonald To be fair, yours was particularly clever hehe... Anyways, unless the rules change to allow modulo (which it totally should, given how ubiquitous it is), I've added a note to my answer indicating that it's not technically valid. Thanks again everyone =)
$endgroup$
– Mwr247
Apr 14 '15 at 18:40
1
1
$begingroup$
@jjdb Parentheses are used to override the normal order of operations, and themselves have a place in that order (albeit at the top). When not specified, the implicit order is to solve from left to right. Parentheses are therefore not required, but rather, recommended for clarity. The case you gave as an example, without parenthesis, would evaluate as: 78 / 10 / 7 = 1.1143... approx.
$endgroup$
– Mwr247
Apr 16 '15 at 13:37
$begingroup$
@jjdb Parentheses are used to override the normal order of operations, and themselves have a place in that order (albeit at the top). When not specified, the implicit order is to solve from left to right. Parentheses are therefore not required, but rather, recommended for clarity. The case you gave as an example, without parenthesis, would evaluate as: 78 / 10 / 7 = 1.1143... approx.
$endgroup$
– Mwr247
Apr 16 '15 at 13:37
|
show 6 more comments
$begingroup$
Addendum too long for a comment
I can confirm there is no better solution than Andrew Smith's if we use the operators +-*/.^ There is only one solution using 3 operators. There are two using 4:
123 +45 - 67 + 8 - 9
123 + 4- 5 + 67 - 89
There are 215 possibilities total and the number of solutions for the number of operators breaks down as follows:
Operators Solutions
0 0
1 0
2 0
3 1
4 2
5 9
6 57
7 104
8 42
We could throw in some more operators though I doubt they would help. I didn't try parentheses, floors, mod, sqrt, etc. If anyone wants a go, though, here's some (unoptimized) VBA that will brute force the basic operators in Excel:
Sub doTheMath()
Const op As String = " +-*/.^"
Dim s As String
Dim a, b, c, d, e, f, g, h
Dim i, j
Dim v
'Loop all possibilities
i = 2
j = 1
For a = 1 To Len(op): For b = 1 To Len(op): For c = 1 To Len(op): For d = 1 To Len(op)
For e = 1 To Len(op): For f = 1 To Len(op): For g = 1 To Len(op): For h = 1 To Len(op)
s = 1 & Mid(op, a, 1) & 2 & Mid(op, b, 1) & 3 & Mid(op, c, 1) & 4 & Mid(op, d, 1) & 5 & _
Mid(op, e, 1) & 6 & Mid(op, f, 1) & 7 & Mid(op, g, 1) & 8 & Mid(op, h, 1) & 9
v = Application.Evaluate(Replace(s, " ", ""))
If CStr(v) = CStr(100) Then
Cells(i, 1).Value = s
Cells(i, 2).Value = v
i = i + 1
End If
j = j + 1
If j > 1000000 Then
ThisWorkbook.Save
j = 1
End If
Next: Next: Next: Next: Next: Next: Next: Next
End Sub
answered Apr 13 '15 at 20:34
Engineer ToastEngineer Toast
11.7k136131
$endgroup$
add a comment |
$begingroup$
Addendum too long for a comment
I can confirm there is no better solution than Andrew Smith's if we use the operators +-*/.^ There is only one solution using 3 operators. There are two using 4:
123 +45 - 67 + 8 - 9
123 + 4- 5 + 67 - 89
There are 215 possibilities total and the number of solutions for the number of operators breaks down as follows:
Operators Solutions
0 0
1 0
2 0
3 1
4 2
5 9
6 57
7 104
8 42
We could throw in some more operators though I doubt they would help. I didn't try parentheses, floors, mod, sqrt, etc. If anyone wants a go, though, here's some (unoptimized) VBA that will brute force the basic operators in Excel:
Sub doTheMath()
Const op As String = " +-*/.^"
Dim s As String
Dim a, b, c, d, e, f, g, h
Dim i, j
Dim v
'Loop all possibilities
i = 2
j = 1
For a = 1 To Len(op): For b = 1 To Len(op): For c = 1 To Len(op): For d = 1 To Len(op)
For e = 1 To Len(op): For f = 1 To Len(op): For g = 1 To Len(op): For h = 1 To Len(op)
s = 1 & Mid(op, a, 1) & 2 & Mid(op, b, 1) & 3 & Mid(op, c, 1) & 4 & Mid(op, d, 1) & 5 & _
Mid(op, e, 1) & 6 & Mid(op, f, 1) & 7 & Mid(op, g, 1) & 8 & Mid(op, h, 1) & 9
v = Application.Evaluate(Replace(s, " ", ""))
If CStr(v) = CStr(100) Then
Cells(i, 1).Value = s
Cells(i, 2).Value = v
i = i + 1
End If
j = j + 1
If j > 1000000 Then
ThisWorkbook.Save
j = 1
End If
Next: Next: Next: Next: Next: Next: Next: Next
End Sub
answered Apr 13 '15 at 20:34
Engineer ToastEngineer Toast
11.7k136131
$endgroup$
add a comment |
$begingroup$
Addendum too long for a comment
I can confirm there is no better solution than Andrew Smith's if we use the operators +-*/.^ There is only one solution using 3 operators. There are two using 4:
123 +45 - 67 + 8 - 9
123 + 4- 5 + 67 - 89
There are 215 possibilities total and the number of solutions for the number of operators breaks down as follows:
Operators Solutions
0 0
1 0
2 0
3 1
4 2
5 9
6 57
7 104
8 42
We could throw in some more operators though I doubt they would help. I didn't try parentheses, floors, mod, sqrt, etc. If anyone wants a go, though, here's some (unoptimized) VBA that will brute force the basic operators in Excel:
Sub doTheMath()
Const op As String = " +-*/.^"
Dim s As String
Dim a, b, c, d, e, f, g, h
Dim i, j
Dim v
'Loop all possibilities
i = 2
j = 1
For a = 1 To Len(op): For b = 1 To Len(op): For c = 1 To Len(op): For d = 1 To Len(op)
For e = 1 To Len(op): For f = 1 To Len(op): For g = 1 To Len(op): For h = 1 To Len(op)
s = 1 & Mid(op, a, 1) & 2 & Mid(op, b, 1) & 3 & Mid(op, c, 1) & 4 & Mid(op, d, 1) & 5 & _
Mid(op, e, 1) & 6 & Mid(op, f, 1) & 7 & Mid(op, g, 1) & 8 & Mid(op, h, 1) & 9
v = Application.Evaluate(Replace(s, " ", ""))
If CStr(v) = CStr(100) Then
Cells(i, 1).Value = s
Cells(i, 2).Value = v
i = i + 1
End If
j = j + 1
If j > 1000000 Then
ThisWorkbook.Save
j = 1
End If
Next: Next: Next: Next: Next: Next: Next: Next
End Sub
answered Apr 13 '15 at 20:34
Engineer ToastEngineer Toast
11.7k136131
$endgroup$
Addendum too long for a comment
I can confirm there is no better solution than Andrew Smith's if we use the operators +-*/.^ There is only one solution using 3 operators. There are two using 4:
123 +45 - 67 + 8 - 9
123 + 4- 5 + 67 - 89
There are 215 possibilities total and the number of solutions for the number of operators breaks down as follows:
Operators Solutions
0 0
1 0
2 0
3 1
4 2
5 9
6 57
7 104
8 42
We could throw in some more operators though I doubt they would help. I didn't try parentheses, floors, mod, sqrt, etc. If anyone wants a go, though, here's some (unoptimized) VBA that will brute force the basic operators in Excel:
Sub doTheMath()
Const op As String = " +-*/.^"
Dim s As String
Dim a, b, c, d, e, f, g, h
Dim i, j
Dim v
'Loop all possibilities
i = 2
j = 1
For a = 1 To Len(op): For b = 1 To Len(op): For c = 1 To Len(op): For d = 1 To Len(op)
For e = 1 To Len(op): For f = 1 To Len(op): For g = 1 To Len(op): For h = 1 To Len(op)
s = 1 & Mid(op, a, 1) & 2 & Mid(op, b, 1) & 3 & Mid(op, c, 1) & 4 & Mid(op, d, 1) & 5 & _
Mid(op, e, 1) & 6 & Mid(op, f, 1) & 7 & Mid(op, g, 1) & 8 & Mid(op, h, 1) & 9
v = Application.Evaluate(Replace(s, " ", ""))
If CStr(v) = CStr(100) Then
Cells(i, 1).Value = s
Cells(i, 2).Value = v
i = i + 1
End If
j = j + 1
If j > 1000000 Then
ThisWorkbook.Save
j = 1
End If
Next: Next: Next: Next: Next: Next: Next: Next
End Sub
answered Apr 13 '15 at 20:34
Engineer ToastEngineer Toast
11.7k136131
answered Apr 13 '15 at 20:34
Engineer ToastEngineer Toast
11.7k136131
answered Apr 13 '15 at 20:34
Engineer ToastEngineer Toast
11.7k136131
answered Apr 13 '15 at 20:34
Engineer ToastEngineer Toast
11.7k136131
11.7k136131
add a comment |
add a comment |
$begingroup$
This is my 2nd attempt.
This time I'm using 7 operations.
I couldn't find any shorter than this, for now.
-(12+34-56)*(-7+8+9)
I've tried and tried and tried.
This simply multiplies 10 by 10.
Nothing fancy, sadly too long, nothing that can't be done easily...
But well, it solves the case!
answered Apr 14 '15 at 19:10
Ismael MiguelIsmael Miguel
302212
$endgroup$
add a comment |
$begingroup$
This is my 2nd attempt.
This time I'm using 7 operations.
I couldn't find any shorter than this, for now.
-(12+34-56)*(-7+8+9)
I've tried and tried and tried.
This simply multiplies 10 by 10.
Nothing fancy, sadly too long, nothing that can't be done easily...
But well, it solves the case!
answered Apr 14 '15 at 19:10
Ismael MiguelIsmael Miguel
302212
$endgroup$
add a comment |
$begingroup$
This is my 2nd attempt.
This time I'm using 7 operations.
I couldn't find any shorter than this, for now.
-(12+34-56)*(-7+8+9)
I've tried and tried and tried.
This simply multiplies 10 by 10.
Nothing fancy, sadly too long, nothing that can't be done easily...
But well, it solves the case!
answered Apr 14 '15 at 19:10
Ismael MiguelIsmael Miguel
302212
$endgroup$
This is my 2nd attempt.
This time I'm using 7 operations.
I couldn't find any shorter than this, for now.
-(12+34-56)*(-7+8+9)
I've tried and tried and tried.
This simply multiplies 10 by 10.
Nothing fancy, sadly too long, nothing that can't be done easily...
But well, it solves the case!
answered Apr 14 '15 at 19:10
Ismael MiguelIsmael Miguel
302212
edited Apr 15 '15 at 9:31
answered Apr 14 '15 at 19:10
Ismael MiguelIsmael Miguel
302212
answered Apr 14 '15 at 19:10
Ismael MiguelIsmael Miguel
302212
answered Apr 14 '15 at 19:10
Ismael MiguelIsmael Miguel
302212
302212
add a comment |
add a comment |
$begingroup$
I know that technically I do not have a shorter solution, but I applied the modulo operator and found an answer!
123456789 % 123456689 = 100
and so on. The program I used:
#include <iostream>
int
main()
{
for (int i = 100; i < 999999999; i++)
{
if (123456789 % i == 100)
{
std::cout << i << std::endl;
}
}
}
answered Apr 14 '15 at 11:22
TheolodisTheolodis
1254
$endgroup$
6
$begingroup$
Welcome to Puzzling SE! Your answer seems to be a bit off-topic, you aren't supposed to use additional numbers than those given by the author!
$endgroup$
– leoll2
Apr 14 '15 at 11:38
$begingroup$
@leoll2 oops, I did read "what is the fewest number or operators"
$endgroup$
– Theolodis
Apr 14 '15 at 14:06
$begingroup$
of, not or. Don't worry, it happens :)
$endgroup$
– Joe Z.
Apr 14 '15 at 16:11
$begingroup$
123456789 % 169 = 92, not 100.
$endgroup$
– Engineer Toast
Apr 18 '18 at 13:48
add a comment |
$begingroup$
I know that technically I do not have a shorter solution, but I applied the modulo operator and found an answer!
123456789 % 123456689 = 100
and so on. The program I used:
#include <iostream>
int
main()
{
for (int i = 100; i < 999999999; i++)
{
if (123456789 % i == 100)
{
std::cout << i << std::endl;
}
}
}
answered Apr 14 '15 at 11:22
TheolodisTheolodis
1254
$endgroup$
6
$begingroup$
Welcome to Puzzling SE! Your answer seems to be a bit off-topic, you aren't supposed to use additional numbers than those given by the author!
$endgroup$
– leoll2
Apr 14 '15 at 11:38
$begingroup$
@leoll2 oops, I did read "what is the fewest number or operators"
$endgroup$
– Theolodis
Apr 14 '15 at 14:06
$begingroup$
of, not or. Don't worry, it happens :)
$endgroup$
– Joe Z.
Apr 14 '15 at 16:11
$begingroup$
123456789 % 169 = 92, not 100.
$endgroup$
– Engineer Toast
Apr 18 '18 at 13:48
add a comment |
$begingroup$
I know that technically I do not have a shorter solution, but I applied the modulo operator and found an answer!
123456789 % 123456689 = 100
and so on. The program I used:
#include <iostream>
int
main()
{
for (int i = 100; i < 999999999; i++)
{
if (123456789 % i == 100)
{
std::cout << i << std::endl;
}
}
}
answered Apr 14 '15 at 11:22
TheolodisTheolodis
1254
$endgroup$
I know that technically I do not have a shorter solution, but I applied the modulo operator and found an answer!
123456789 % 123456689 = 100
and so on. The program I used:
#include <iostream>
int
main()
{
for (int i = 100; i < 999999999; i++)
{
if (123456789 % i == 100)
{
std::cout << i << std::endl;
}
}
}
answered Apr 14 '15 at 11:22
TheolodisTheolodis
1254
edited Apr 20 '18 at 9:23
answered Apr 14 '15 at 11:22
TheolodisTheolodis
1254
answered Apr 14 '15 at 11:22
TheolodisTheolodis
1254
answered Apr 14 '15 at 11:22
TheolodisTheolodis
1254
1254
6
$begingroup$
Welcome to Puzzling SE! Your answer seems to be a bit off-topic, you aren't supposed to use additional numbers than those given by the author!
$endgroup$
– leoll2
Apr 14 '15 at 11:38
$begingroup$
@leoll2 oops, I did read "what is the fewest number or operators"
$endgroup$
– Theolodis
Apr 14 '15 at 14:06
$begingroup$
of, not or. Don't worry, it happens :)
$endgroup$
– Joe Z.
Apr 14 '15 at 16:11
$begingroup$
123456789 % 169 = 92, not 100.
$endgroup$
– Engineer Toast
Apr 18 '18 at 13:48
add a comment |
6
$begingroup$
Welcome to Puzzling SE! Your answer seems to be a bit off-topic, you aren't supposed to use additional numbers than those given by the author!
$endgroup$
– leoll2
Apr 14 '15 at 11:38
$begingroup$
@leoll2 oops, I did read "what is the fewest number or operators"
$endgroup$
– Theolodis
Apr 14 '15 at 14:06
$begingroup$
of, not or. Don't worry, it happens :)
$endgroup$
– Joe Z.
Apr 14 '15 at 16:11
$begingroup$
123456789 % 169 = 92, not 100.
$endgroup$
– Engineer Toast
Apr 18 '18 at 13:48
6
6
$begingroup$
Welcome to Puzzling SE! Your answer seems to be a bit off-topic, you aren't supposed to use additional numbers than those given by the author!
$endgroup$
– leoll2
Apr 14 '15 at 11:38
$begingroup$
Welcome to Puzzling SE! Your answer seems to be a bit off-topic, you aren't supposed to use additional numbers than those given by the author!
$endgroup$
– leoll2
Apr 14 '15 at 11:38
$begingroup$
@leoll2 oops, I did read "what is the fewest number or operators"
$endgroup$
– Theolodis
Apr 14 '15 at 14:06
$begingroup$
@leoll2 oops, I did read "what is the fewest number or operators"
$endgroup$
– Theolodis
Apr 14 '15 at 14:06
$begingroup$
of, not or. Don't worry, it happens :)
$endgroup$
– Joe Z.
Apr 14 '15 at 16:11
$begingroup$
of, not or. Don't worry, it happens :)
$endgroup$
– Joe Z.
Apr 14 '15 at 16:11
$begingroup$
123456789 % 169 = 92, not 100.
$endgroup$
– Engineer Toast
Apr 18 '18 at 13:48
$begingroup$
123456789 % 169 = 92, not 100.
$endgroup$
– Engineer Toast
Apr 18 '18 at 13:48
add a comment |
$begingroup$
I'm not sure this is what you're looking for, but 123456789=/=100
This answer adds two symbols, "=" and "/".
answered Apr 14 '15 at 0:18
DiscOHDiscOH
44129
$endgroup$
2
$begingroup$
You can't do that, though. The symbol=/=doesn't mean anything, and even if you were trying to change it to $neq$, that's specifically disallowed.
$endgroup$
– Joe Z.
Apr 14 '15 at 0:20
$begingroup$
Would you be happier with "123456789 != 100"? ! meaning "not" is listed in wikipedia's big list of math symbols en.wikipedia.org/wiki/List_of_mathematical_symbols
$endgroup$
– DiscOH
Apr 14 '15 at 1:03
10
$begingroup$
Well, the problem is with the word "equation". Sure you've made the statement true, but it's no longer an equation.
$endgroup$
– Joe Z.
Apr 14 '15 at 1:37
$begingroup$
@JoeZ. An equation is an equality with at least one variable. What you mean is an equality.
$endgroup$
– user26486
Apr 14 '15 at 23:23
$begingroup$
Well, in any case I've edited the question statement again and that removes any ambiguity with regards to my choice of word when it comes to "equation".
$endgroup$
– Joe Z.
Apr 15 '15 at 0:04
add a comment |
$begingroup$
I'm not sure this is what you're looking for, but 123456789=/=100
This answer adds two symbols, "=" and "/".
answered Apr 14 '15 at 0:18
DiscOHDiscOH
44129
$endgroup$
2
$begingroup$
You can't do that, though. The symbol=/=doesn't mean anything, and even if you were trying to change it to $neq$, that's specifically disallowed.
$endgroup$
– Joe Z.
Apr 14 '15 at 0:20
$begingroup$
Would you be happier with "123456789 != 100"? ! meaning "not" is listed in wikipedia's big list of math symbols en.wikipedia.org/wiki/List_of_mathematical_symbols
$endgroup$
– DiscOH
Apr 14 '15 at 1:03
10
$begingroup$
Well, the problem is with the word "equation". Sure you've made the statement true, but it's no longer an equation.
$endgroup$
– Joe Z.
Apr 14 '15 at 1:37
$begingroup$
@JoeZ. An equation is an equality with at least one variable. What you mean is an equality.
$endgroup$
– user26486
Apr 14 '15 at 23:23
$begingroup$
Well, in any case I've edited the question statement again and that removes any ambiguity with regards to my choice of word when it comes to "equation".
$endgroup$
– Joe Z.
Apr 15 '15 at 0:04
add a comment |
$begingroup$
I'm not sure this is what you're looking for, but 123456789=/=100
This answer adds two symbols, "=" and "/".
answered Apr 14 '15 at 0:18
DiscOHDiscOH
44129
$endgroup$
I'm not sure this is what you're looking for, but 123456789=/=100
This answer adds two symbols, "=" and "/".
answered Apr 14 '15 at 0:18
DiscOHDiscOH
44129
answered Apr 14 '15 at 0:18
DiscOHDiscOH
44129
answered Apr 14 '15 at 0:18
DiscOHDiscOH
44129
answered Apr 14 '15 at 0:18
DiscOHDiscOH
44129
44129
2
$begingroup$
You can't do that, though. The symbol=/=doesn't mean anything, and even if you were trying to change it to $neq$, that's specifically disallowed.
$endgroup$
– Joe Z.
Apr 14 '15 at 0:20
$begingroup$
Would you be happier with "123456789 != 100"? ! meaning "not" is listed in wikipedia's big list of math symbols en.wikipedia.org/wiki/List_of_mathematical_symbols
$endgroup$
– DiscOH
Apr 14 '15 at 1:03
10
$begingroup$
Well, the problem is with the word "equation". Sure you've made the statement true, but it's no longer an equation.
$endgroup$
– Joe Z.
Apr 14 '15 at 1:37
$begingroup$
@JoeZ. An equation is an equality with at least one variable. What you mean is an equality.
$endgroup$
– user26486
Apr 14 '15 at 23:23
$begingroup$
Well, in any case I've edited the question statement again and that removes any ambiguity with regards to my choice of word when it comes to "equation".
$endgroup$
– Joe Z.
Apr 15 '15 at 0:04
add a comment |
2
$begingroup$
You can't do that, though. The symbol=/=doesn't mean anything, and even if you were trying to change it to $neq$, that's specifically disallowed.
$endgroup$
– Joe Z.
Apr 14 '15 at 0:20
$begingroup$
Would you be happier with "123456789 != 100"? ! meaning "not" is listed in wikipedia's big list of math symbols en.wikipedia.org/wiki/List_of_mathematical_symbols
$endgroup$
– DiscOH
Apr 14 '15 at 1:03
10
$begingroup$
Well, the problem is with the word "equation". Sure you've made the statement true, but it's no longer an equation.
$endgroup$
– Joe Z.
Apr 14 '15 at 1:37
$begingroup$
@JoeZ. An equation is an equality with at least one variable. What you mean is an equality.
$endgroup$
– user26486
Apr 14 '15 at 23:23
$begingroup$
Well, in any case I've edited the question statement again and that removes any ambiguity with regards to my choice of word when it comes to "equation".
$endgroup$
– Joe Z.
Apr 15 '15 at 0:04
2
2
$begingroup$
You can't do that, though. The symbol
=/= doesn't mean anything, and even if you were trying to change it to $neq$, that's specifically disallowed.$endgroup$
– Joe Z.
Apr 14 '15 at 0:20
$begingroup$
You can't do that, though. The symbol
=/= doesn't mean anything, and even if you were trying to change it to $neq$, that's specifically disallowed.$endgroup$
– Joe Z.
Apr 14 '15 at 0:20
$begingroup$
Would you be happier with "123456789 != 100"? ! meaning "not" is listed in wikipedia's big list of math symbols en.wikipedia.org/wiki/List_of_mathematical_symbols
$endgroup$
– DiscOH
Apr 14 '15 at 1:03
$begingroup$
Would you be happier with "123456789 != 100"? ! meaning "not" is listed in wikipedia's big list of math symbols en.wikipedia.org/wiki/List_of_mathematical_symbols
$endgroup$
– DiscOH
Apr 14 '15 at 1:03
10
10
$begingroup$
Well, the problem is with the word "equation". Sure you've made the statement true, but it's no longer an equation.
$endgroup$
– Joe Z.
Apr 14 '15 at 1:37
$begingroup$
Well, the problem is with the word "equation". Sure you've made the statement true, but it's no longer an equation.
$endgroup$
– Joe Z.
Apr 14 '15 at 1:37
$begingroup$
@JoeZ. An equation is an equality with at least one variable. What you mean is an equality.
$endgroup$
– user26486
Apr 14 '15 at 23:23
$begingroup$
@JoeZ. An equation is an equality with at least one variable. What you mean is an equality.
$endgroup$
– user26486
Apr 14 '15 at 23:23
$begingroup$
Well, in any case I've edited the question statement again and that removes any ambiguity with regards to my choice of word when it comes to "equation".
$endgroup$
– Joe Z.
Apr 15 '15 at 0:04
$begingroup$
Well, in any case I've edited the question statement again and that removes any ambiguity with regards to my choice of word when it comes to "equation".
$endgroup$
– Joe Z.
Apr 15 '15 at 0:04
add a comment |
protected by Community♦ Apr 14 '15 at 19:00
Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead?
4
$begingroup$
Can I insert a symbol between = and 100?
$endgroup$
– Ben Frankel
Apr 13 '15 at 19:45
10
$begingroup$
@JoeZ. You should define the question such that you don't need to make ex post facto judgements on what is or isn't allowed. You failed to specify specifically which "mathematical symbols" are allowed and their meaning. Numbers are symbols, and they are used in mathematics. Taking it a step farther, I could invent a new operator with its own symbol (as is often done in mathematics) and define it to always evaluate to 100. Stick it anywhere on the left and voila: 100 = 100, one added symbol.
$endgroup$
– bcrist
Apr 14 '15 at 1:10
3
$begingroup$
@JoeZ. I wasn't complaining or implying that my hypothetical solution should be considered a valid solution to the puzzle, but trying to point out that if you explicitly listed the specific operators and their acceptable meanings (some operators have different or multiple meanings in different areas of mathematics) you would not need to contend with a plethora of "clever" solutions utilizing lateral thinking.
$endgroup$
– bcrist
Apr 14 '15 at 1:47
3
$begingroup$
@JoeZ. I believe my first comment was and still is quite constructive: it gives specific advice for how to describe the puzzle less ambiguously. Namely, by listing each symbol/operator which is acceptable and what meaning it has. What irritates me is when people misinterpret constructive criticism and cry foul instead.
$endgroup$
– bcrist
Apr 14 '15 at 2:07
2
$begingroup$
@bcrist I apologize. I got antsy about this because I've been accused of moving the goalposts in much the same way at PCG.SE (the code golf site) more than once, and it pattern-matches to people complaining about "chameleon questions" where people do indeed believe their loophole-abusing solutions should be considered valid even if they ruin the challenge, almost as a penalty to the question-asker for not making the question clear enough in the first place. Looking back at my first reply to you, some of my negativity was pretty uncalled for, so I've just deleted my comments.
$endgroup$
– Joe Z.
Apr 14 '15 at 5:50