this is driving me nuts...

enter image description here

Any ideas? Been on it for 2 hours now. The only info given, was this:

The solution is not, what you think it is.

I guess it is doable with just integers, but every attempt I tried leads eventually to a dead end. Ideas or solutions?

  • 2
    $\begingroup$ Assuming that this is just math, not lateral-thinking: this is a system of 4 equations with 4 unknowns. You can represent it with a matrix to find a solution to most problems like this, but there is no solution for this particular system. Ie. even with non-integers, you don't have numbers that satisfy these equations. $\endgroup$
    – Lacklub
    Apr 4, 2016 at 13:14
  • 1
    $\begingroup$ You can't solve it with a system (no solution) so it might be lateral thinking $\endgroup$
    – Fabich
    Apr 4, 2016 at 13:18
  • 1
    $\begingroup$ @Lordofdark It's not possible to solve in any other base either, so it's not that type of lateral thinking $\endgroup$
    – Lacklub
    Apr 4, 2016 at 13:19
  • $\begingroup$ What is the format of the solution ? 4 integers ? 4 real numbers ? $\endgroup$
    – Fabich
    Apr 4, 2016 at 13:21
  • 1
    $\begingroup$ I am not english native, isn't the "," in the hint a little strange ? $\endgroup$
    – Fabich
    Apr 4, 2016 at 13:35

2 Answers 2


If you think of this simply as math, then the answer is:

No solution exists.

Let the blue square be A, the red square be B, the yellow C and the green D. Then the following equations are required:

$A - B = 9$ (1)

$C - D = 14$ (2)

$A + C = 12$ (3)

$B + D = 2$ (4)

If you add (1) and (2), you get the equation:

$$A - B + C - D = 9 + 14$$

$$A + C - (B + D) = 23$$

If you then add equation 4:

$$A + C - (B + D) + (B + D) = 23 + 2 $$

$$A + C = 25$$

Which obviously contradicts equation (3). Therefore, it is impossible to find any real numbers that satisfy all of the equations.

This might actually be the correct answer to the question, with no lateral thinking required: the hint says "the solution is not" in an odd way, which could be hinting at the non-existence of the solution (it is not = it doesn't exist). Thanks to Lord of dark for the idea.

  • $\begingroup$ Neat idea actually... If you put it in this context: "The answer is not. What do you think it is?", it would make perfect sense.. wouldn't it? $\endgroup$
    – DasSaffe
    Apr 4, 2016 at 13:45
  • $\begingroup$ @DasSaffe Exactly. It would also make a lot of sense if the question was "Can you find four numbers that solve the problem?". Then the answer is that you can not find those numbers. The answer is not. $\endgroup$
    – Lacklub
    Apr 4, 2016 at 13:48
  • 2
    $\begingroup$ I think I've just understand the hint: "The solution is not, what you think it is". It's not a single hint but dual: "The solution is not". It told us. "What you think it is" it's in fact an other hint. You think this equations can't be solvable? You're right. $\endgroup$
    – Shkeil
    Apr 4, 2016 at 13:48
  • 1
    $\begingroup$ should you be adding 2 instead of 4 when adding equation 4? $\endgroup$
    – Bishop
    Apr 4, 2016 at 14:27

Although there is a correct answer marked, I want to come with a different approach and point of view.

1: A−B=9

2: C−D=14

3: A+C=12

4: B+D=2

By adding 2 & 4, we get B+C = 16

By subtracting 1 from 3, we get B+C = 3

Now tricky part. Consider an imaginary hidden absolute value for C on second result, then the numbers are:

B =  9.5
C = -6.5

From equation 1 & 4 we get other two as well so now we have:

A = 18.5
B =  9.5
C = -6.5
D = -7.5

Equations 1,3,4 fits perfectly with these numbers.

To fit equation 2; we can assume a "hidden" rule which does not effect other equations but this; "if an equation starts with a negative number, take absolute of that number". Only equation suits this info is #2.

Since this is a puzzle, and there is no rule indicating that there can't be additional or hidden rules, these numbers with the given rule fits as an answer.

The solution is not, what you think it is. True, there were hidden things in the puzzle

  • 5
    $\begingroup$ It's not hidden, you made up a rule to fit equation 2. I would consider this a standard loophole $\endgroup$
    – Narmer
    Apr 4, 2016 at 15:30

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