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The puzzle: enter image description here

The solution is a single number, which replaces the question mark. Good luck!

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    $\begingroup$ I'm not very happy about the mystery-meat link there. For all we know it could go to a pornographic website, or something serving up malware, or a commercial site selling recreational pharmaceuticals, etc. I'm guessing from the fbcdn.net that it's actually one of those "man-with-hat plus tweety-bird plus thistle = 26 ..." puzzles that circulate on Facebook, but who knows? Would you consider either finding a suitable image you can legally upload to stack.imgur.com, or giving some sort of description instead of a link? $\endgroup$ – Gareth McCaughan Apr 9 '20 at 21:30
  • $\begingroup$ @GarethMcCaughan - I took care of the image. $\endgroup$ – Lanny Strack Apr 9 '20 at 21:57
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Original Answer:

The trick here is,

! in the puzzle is used in two ways, both as a variable and as the sign of factorial!

So,

The solution is $\text{?}=\text{!}=2$.

And,

The three equations are actually:
$\frac{2+2}{2}=\frac{2\times 2 \times 2}{2 \times 2!}$
$2 \times 2-2!!=\frac{2 \times 2}{2}$
$2=2$


After reading @Firecase's explanation,

Actually the last line is just asking for the value of $\text{!}$, like $x=\text{?}$ in a normal math question.

So we can get another solution:

$\text{?}=4, \text{!}=2$.
$\frac{4+2}{2}=\frac{2\times 4!}{2 \times 4 \times 2}$
$4 \times 4-2!\times 2=\frac{4!}{2}$

I believe there're no other integer solutions.

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  • $\begingroup$ ? doesn't equal to ! Semantically, that ? is asking the solution itself (As asked in the question. Its role is to take up the value of !, but only once). Just like "?", "!" have two different roles (That's the part you discovered.) ! is indeed 2, and it`s the correct answer, but there is different value for "?" maybe you can investigate it. However I accept this answer because there is ambiguity in the question that can lead someone to this "almost correct" answer. $\endgroup$ – FIreCase Apr 10 '20 at 10:10
  • $\begingroup$ I added the other solution, is it correct? @FIreCase $\endgroup$ – newbie Apr 10 '20 at 10:49
  • $\begingroup$ Totally correct. I investigated quite a bit beforehand to exclude other solutions (mainly among higher values, seemingly I didn't manage to, because you simply equated the two), but now I believe there are no more pairs. $\endgroup$ – FIreCase Apr 10 '20 at 10:56
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Puzzle

Let's $(x,y) \in \mathbb R_*^2$

$\text{Using equations }1 \text{ and }3$

$\dfrac{x+y}y =1$ and $x = y$ $\implies\text{there is no solution for our set of three equations}$

Image

1. $x$ is number (or value) of a pair of shoes
2. $y$ is number (or value) of a guy
3. $z$ is number (or value) of a sunglass
4. $w$ is number (or value) of a glove

$3x = 60 \implies x = 20$
$20 + 2y= 30 \implies y = 5$
$2z + y = 9 \implies z = 2$
$20+w+ 2 = 42 \implies w = 20$
$? = \dfrac{x}2 + (x + y + 2w + z) \times z= 10 + 67\times2 = 144$

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IMAGE

6x = 60 => x = 10

2x + 2y = 30 => y = 5

2z + y = 9 => z = 2

2x + w + z = 42 => w = 20

? = x + [(y + 2x + 2w + z) * z => ? = x + (67 * 2) =>

? = 144

EQUATIONS

The equations are valid only for ? = 0 and ! = 0

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    $\begingroup$ I don't think this answers the question, as it states the picture isn't the puzzle for solving here $\endgroup$ – Beastly Gerbil Apr 9 '20 at 23:04
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    $\begingroup$ That was not clear from the beginning Beastly. The question has been modified. $\endgroup$ – Stratos Apr 9 '20 at 23:07
  • $\begingroup$ If both ! and ? are 0, you're dividing by 0, so there is no solution as explained in JKHA's answer. $\endgroup$ – mbjb Apr 10 '20 at 8:09
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My quess:

2

Reasoning/thoughts:

I the left part of the first equation and the third equation are taken into account, the following holds: $\frac{?+!}{!} = \frac{!+!}{!} = \frac{2!}{!} = 2$. Thus now the solution is already found, the right side of the first equation and the second equation aren't needed.

For the sake of completeness the redundant equations are investigated. First I want to note that there are two options for the meaning of multiple question and/or exclamation marks after each other (for example: ?!?): there are two options:

1. !?! means $!*?*!$, then $\frac{!*?*!}{!*?*!} = 1$.

2. !?! means $!?!$ writing it as a 3-digit number. Assuming $! = ? = 2$ then $!?!$ would be $222$ and $\frac{!?!}{!?!} = 1$ (regardless of the number for ?).

So both options would conclude that the first equation is mathematically not possible... Looking at the second equation with the same two options:

1. $?? - !!! = \frac{?!}{!} $ means $?*? - !*!*! = \frac{?*!}{!} = ?$. Filling the third equation in: $2*? - 3*? = ?$, which is mathematically incorrect (should be $-?$).

2. $?? - !!! = \frac{?!}{!} $ will never be mathematically correct, because a 3-digit number is subtracted from a 2-digit number, thus always negative and $\frac{?!}{!} $ could never be negative. I'm not sure how this option would work with negative numbers: for example $?? = -2-2$ or something?

So I don't think there is an answer, but my guess would be 2 based on the left part op equation 1 and eaquation 3.

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