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I made this simple puzzle a couple of years ago. There are three colors which appear in every row and every column. Moreover, every color appears 5 or 6 times in total. Find the color of the square with the question mark.

enter image description here

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    $\begingroup$ Grey. Everything is grey when you are colorblind. $\endgroup$ – Marius Feb 23 '17 at 9:34
  • $\begingroup$ @Marius false - achromatopsia is pretty rare. Deuteranoptics and tritanoptics won't have issues with this challenge (except for reporting the results). Protanoptics (the most common form) will have a harder time. In any case: the left column has a green tile filled in, the second column has two red, the right two columns have blue. $\endgroup$ – John Dvorak Feb 24 '17 at 2:38
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    $\begingroup$ It's your checkmark to award, but - why isn't boboquack's answer the Accepted one? $\endgroup$ – Rubio Feb 24 '17 at 3:00
  • $\begingroup$ @Rubio At first I am new here and maybe I am misinterpreting something. Even though I enjoyed reading boboquack's answer, JonMark's answer is shorter and seems more elegant to me. $\endgroup$ – Alexandros Feb 24 '17 at 3:50
  • $\begingroup$ It's predominantly the case that of equivalent answers, the earlier one gets Accepted; of not equivalent answers, the one with more detail that would be of greater use to other solvers (the why, not merely the answer) wins out. On both counts, bobo's seems to me to be the better answer to accept. $\endgroup$ – Rubio Feb 24 '17 at 3:54
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I get:

44block Top right is fairly obvious, the 3rd row needs a red and green, and the red can't go in column 2.

Top row and last column now need red/green, and first column, last row need a blue, so the blues go.

Now we have 6 blues, and so we must have 5 reds and 5 greens. As top row/last column both need red/green, this gives us 5 reds, so ? is green.

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  • $\begingroup$ Just a note that, while the puzzle doesn't strictly require it, you could fill in one more red in the third column, giving you one more green in the first column. Although that still leaves two squares that I think have to remain ambiguous. $\endgroup$ – Admiral Jota Feb 23 '17 at 23:40
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Answer for the question mark:

Green

Solution:

Actually, there are two!
Answer 1
or
Answer 2

Working:

If we look at the right column, we can't get any more blues otherwise the red and green couldn't fit in. These are marked with a diagonal line:
A
Then the bottom row must have blue as shown:
B
We can apply the same argument now to the third column and then the second row:
C
And to the second column and the third row, but in red:
D
So now we can find the green in the third row:
E
And the blue in the second column:
F
We can't have another blue in the top row, so the remaining rows columns apart from the second must each have a red and a green:
G
So there's exactly one green in the third column and exactly one more in the fourth. Since there is at least 5 greens, the top-left square must be green:
H
Because the first row and the third column each need to have one red and one green, we have the following:
I
And then we have the ambiguity of which colours go where, and this cannot be resolved.

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0
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I get

Green

with a finished board of

GBRB
BRGR
RGBB
GRBG
(6 Blue, 5 Red, 5 Green)

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  • 3
    $\begingroup$ Thanks for your answer! You might want to have a look at the other answers too - they have come to the same conclusion! $\endgroup$ – boboquack Feb 23 '17 at 21:26

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