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Can you solve this one? I'll give a second hint if needed.

enter image description here

hint

answer and reasoning can both be found in the title of the question. the correct answer is literally irrefutable according to this reasoning.

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It's:

C

because:

Every row and column contains either a $1:2$ or $2:1$ ratio of squares with an odd:even number of lines. Both the bottom row and right column already have two even squares, so the final square must be an odd square, and C is the only odd square available.

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  • $\begingroup$ This may also be correct but it does not prevent another possibility from being correct, following the same kind of logic. I'll post a third possibility as an answer, and give a hint in the body of the question. $\endgroup$ – qq jkztd Jul 7 at 14:27
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I Think the answer is option

a)

Reason

Along each row and column, the number of lines increases(row) and decreases(column) by $1$. $1^{st}$ row $\implies 4$, $2^{nd}$ row $\implies 5$ and $3^{rd}$ row $\implies 6$. Already there're $6$ lines, so the last box should be empty. Similarly for the columns.

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  • $\begingroup$ it works to some extent, but this arithmetic reasoning does not prevent another possibility from being correct as well. $\endgroup$ – qq jkztd Jul 7 at 13:57
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It could also be

b)

Because

each row and colum has two times the same amount of heavy lines. enter image description here

But still this is not the correct reason.

EDIT: here is how I built it and what I was aiming for. I'll improve tags and narrow down other possible answers for the next one.

enter image description here

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  • $\begingroup$ Self-answers are, of course, perfectly ok here as they are anywhere else on SE. Having said that, it is generally nicer to give solvers some hints and some time to let them come up with the full solution themselves. $\endgroup$ – Rubio Jul 11 at 6:18

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