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I've tried to solve this question(Exercise 31) in many ways but I couldn't figure out the logic behind this.

$\Space{100pt}{0px}{0px}$ enter image description here

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My solution:

D

My reasoning:

Treat the triples as vectors over a set with elements B (black), W (white) and S (striped). The vectors in the first two rows are "added" (with some noncommutative binary operation), the third row is the result. From the first elements in the first column, we get S + S = W. This leaves the solutions B, D and E. B and E can be eliminated because W + S can't be two values at the same time.

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  • $\begingroup$ B + B = W or S (row 1 & 2), so why can't W + S be two values at the same time? $\endgroup$ – zcahfg2 Nov 25 at 19:28
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I think the answer is

'A'. There are 9 white squares - 3 in the right position, 3 in the middle, 3 in the left spot. There are 10 striped - 4 left, 3 middle, 3 right. There are 5 blacks - 1 left, 2 middle, 2 right.
Option A would give you:
6 blacks, 2 at each position.
9 whites, 3 at each position.
12 striped, 4 at each position.

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  • $\begingroup$ Hi @Nicko, welcome to Puzzling SE! Take the tour if you haven't already! This is a great first answer, and I've reformatted it to include spoiler tags using >! in order to hide your solution. I hope this helps! $\endgroup$ – HTM Oct 16 at 23:32
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    $\begingroup$ Thank you, and it does help. I now know how to hide answers and will do so. $\endgroup$ – Nicko Oct 16 at 23:34
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I would say possibly A because then the number of whites are decreasing by -1 if you add the number of white squares per subsequent column from left to right. Also there is another contingent pattern it solves with the sum of blacks per subsequent row from up to down decreasing by -1. No other answer satisfies the first logic anyway. So it is sufficient enough to stand on its own as the sole reason.

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My solution:

D

My reasoning:

The patters works both in rows and columns: The first and third figures in the third row and third columns are white & grey. Therefore, whatever the results is, the first and third figure must be the same, which leaves us with options D and F. From the first row, we know that white & grey is grey, which leaves us D.

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This is not a full solution. It is brainstorming that others might be able to use.. if I come up with a solution that meets all requirements I will delete this comment and append the solution to the bottom.

Let's first look at the solution posted by Alexander Fasching.

- Overlay the left column with the center column.
- Let G be striped (grey), B be black, and W be white.
- Wherever G and B overlap the result is W
- Wherever W and G overlap the result is G
- W and W overlapping is undefined.
* - But if you take the first horizontal row and overlay it with the middle horizontal row you get B+W=W on column 1 and B+W=G on column 2.
* - Another break is to work on the columns left to right, where it breaks because on Row 1 B+B=W but on Row 2 B+B=G

Therefore I am thinking there must be another solution..

That being said, what if we look at Deepthinker101's solution.

- First off, as they pointed out choosing A allows for all small boxes to be grouped into 3 same color sets evenly.
- 3 sets of whites, 2 sets of black, and 4 sets of greys.
- The only issue I have is that it seems arbitrary to break all of the groups apart to meet this goal.. but perhaps there's a pattern that explains it.

So, looking even further into it..

- I found a definite pattern.. - I assigned G as -1, W as 0 and B as 1. I then subtracted C1R1 from C4R2 and arrived at -1 (C9R3). Then I subtracted C1R2 from C4R3 and got C9R1. Then I subtracted C1R3 from C4R1 and got C9R2. You can repeat this system for the entire grid and it works out.. Except it requires a solution of Grey, Grey, Black.
- As this is not an option it can't be the area, but perhaps the pattern can still be used.

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D because horizontally, white + grey -> grey

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    $\begingroup$ Can you expand on this a bit? It's hard to understand what "white + grey -> grey" means. $\endgroup$ – Rand al'Thor Sep 26 at 12:42
  • $\begingroup$ That does not explain why black+black->white on the first row but black+black->grey on the second row. $\endgroup$ – Jaap Scherphuis Sep 26 at 12:49
  • $\begingroup$ b + b can be white or gray. It's fuzzy :D $\endgroup$ – Markus Sep 26 at 13:09
  • $\begingroup$ actually despite of the downvote this is the most straightforward answer for D. (And the intended answer is D because my "IQ" only grew from 95 to 97 when I chose D on question 31 while using the same pattern of answers on the rest of the questions on test.mensa.no/# hehe $\endgroup$ – balazs.com Oct 30 at 14:40

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