Fdzfgy

So remember all those puzzles about 100 or 1000 or however many logicians abducted by aliens and they are going to be killed if they can't figure out their hat color or the number on their back? Well it turns out you were such a bright mind that the aliens abducted you, and luckily you were one of the logicians that survived.

Unfortunately, instead of taking you back home, they just left you in a room with all the other logicians that survived. Escaping doesn't seem very likely any time soon because right now everyone is focusing on getting food every day. There is a single fridge in the corner of the room but it is one of those newfangled fridges and it has a security code that you have to put in to open it. The fridge has a panel on the top that displays 5 digits, and those digits change every day, along with the fridge restocking. The problem is that when the fridge restocks, the security code changes. For the first six days, you've had to brute force the code by just typing every combination in until you get it (only 100000 combinations, not too bad), but you really need to figure out what the pattern is for the codes so you can focus on escaping instead of cheeseburgers.

The panel numbers and codes for that day are as follows:

Panel -- Code

12345 = 56306

04418 = 35129

20530 = 65048

54472 = 68425

00000 = 00000

60454 = 95126

29830 = 39175

00000 = 03334

08224 = 35386

14412 = ?????

Today the panel says 14412. Can you figure out the code and what the algorithm is so you can get the food easily each day?

About the brute-force method: Pertaining to David James' comment,the brute force method started with 00000 on the fifth day, but it does not follow the pattern of 00001, 00002... One logician is in charge of the brute-forcing and each day he starts with the panel number, and then does what seems like a random sequence. At least that's what you all think... he's been a little bit looney since the abduction but you all let him use his "sequence" to make him happy.

Yesterday in the fridge, someone found a fortune cookie which contained the following fortune:

Look backwards to go forward

It's probably just a generic fortune mass produced in a cookie factory, but what if it's a clue about solving the code?

Another fortune cookie has shown up:

651
930
862
565
000
696
532
400

What could those numbers be referring to?

More fortune cookies:

56345
35118
65030
68472
00000
95154
39130
03300
35324

And:

Codes to a day need not connect
But codes to the panel need be
Look how to get panel from code
And you'll be able to get free

Last fortune cookie:

There can be more than one, but each can't be more than once.

Days passed: 9 (seems like a lot more though)

Note: Please post partial answers if you think you have anything so I can see where you are having problems

It's not possible to determine exactly the code to enter on any given day, but we can narrow it down a great deal as follows:

First, let's define the digits of the Panel as being Pn and the digits of the Code as being Cn. Now, there are various mathematical relationships that can be applied, to relate digits of the code to digits of the panel:

• $C2 - C3 = P3$


• $C5 mod C1 = P1$


• $C1(C2 + C3) = P4,P5$


• $C1 - C3 + C4 = P2$

Thus we have a set of equations, which collectively cover all the digits of the code. However, because there's ambiguity (eg. with the mod operator), we still can't directly determine any given code.

Taking the latest panel value of $14412$, we know the following:

• $C2 - C3 = 4$


• $C5 mod C1 = 1$


• $C1(C2 + C3) = 12$


• $C1 - C3 + C4 = 4$

(I'm sure an actual mathematician would do a cleaner job of this part of the logic, but I'll fumble along). We can rework these equations to give us, $C1(4 + 2C3) = 12$ (combining equation 1 & 3), or $C1 = frac{12}{(4 + 2C3)}$.

Given that $C1$ must be a whole number, then $C3$ must be $0$, $1$, $4$, making $C1$ equal to $3$, $2$, or $1$. However, we also need the remainder when dividing $C5$ by $C1$ to be $1$ (noting that there's precedent in the examples for treating x%0 as 0), which means $C1ne1$.

Re-entering these possible values in the third equation to get $C2$, we have our code as either $251??$ or $340??$. Then using the 4th equation to get $C4$, we arrive at $2513?$ or $3401?$. Finally, using the 2nd equation we get a $C5$ value of $1$, $3$, $5$, $7$, $9$ (for the first possible code) and $1$, $4$, $7$ (for the second).

Thus, today's code is one of the following:

• $25131$


• $25133$


• $25135$


• $25137$


• $25139$


• $34011$


• $34014$


• $34017$

So, not perfect, but certainly faster than brute forcing $100,000$ combinations...

14412
75628 28591 62916 48164 91748 58464 74748 28483 81638 18174
74826 26475 83828 49175 74658 37575 75936 36565 81638 17585
75756 46282 92857 46382 75748 38165 81848 56485 64858 56382
72628 36281 81728 16463 75828 16483 63828 58163 63630 47481
91918 46385 84656 48565 62946 26285 91859 17491 72756 46575
71658 36264 74818 28462 82649 18193 65626 48484 91838 57491
81657 27483 83858 28364 62726 26562 83759 27263 82827 27283
82858 47582 81837 28462 82837 58164 75748 58162 92000