8
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I have been creating puzzles to promote our school's Facebook page, but in a week no-one could yet answer this one:

Complete the pattern:

enter image description here

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8
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It is:

C

because:

Let white=0, grey=1, black=2. Adding the first two squares from either rows or columns and subtract 3 if the answer is greater than 2 gives the third square in the row/column.

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5
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Answer is:

C

Because:

Black + black = gray.
Gray + gray = black.
Color + white = color.

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  • 2
    $\begingroup$ What about Black + Gray? $\endgroup$ – ibrahim mahrir Jul 16 '18 at 19:09
  • $\begingroup$ @ibrahimmahrir It looks like Danne used the term "color" to mean either black or grey. That makes sense when also considering how other answers assign the value of $0$ to white, as $x + 0 = x$. $\endgroup$ – maxathousand Jul 16 '18 at 21:03
  • $\begingroup$ Don't forget: White $+$ White $=$ White. $\endgroup$ – Mr Pie Jul 16 '18 at 21:59
  • $\begingroup$ @maxathousand But that when "color" is added to white. I'm asking about Black + Gray. $\endgroup$ – ibrahim mahrir Jul 16 '18 at 22:12
  • $\begingroup$ @ibrahimmahrir oh wow, hahha totally misread that. You’re right $\endgroup$ – maxathousand Jul 17 '18 at 0:38
4
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The answer is:

C.

Due to:

Assign black, grey, and white the values, and 0 respectively (Black would be 2, grey 1, and white 0). Now the top-left square would look like this: $$\begin{bmatrix}2 & 1 & 0\\1 & 2 & 1 \\0 & 1 & 2 \end{bmatrix}$$ And the middle-left like this: $$\begin{bmatrix}0 & 0 & 0\\1 & 1 & 1 \\2 & 2 & 2 \end{bmatrix}$$ Adding them together would result in a square looking like this: $$\begin{bmatrix}2 & 1 & 0\\2 & 3 & 2 \\2 & 3 & 4 \end{bmatrix}$$ Which has values greater than 2, so we use a modulo function (which gives the remainder of a division) to subtract three from all values greater than 2, gives a box with the following values: $$\begin{bmatrix}2 & 1 & 0\\2 & 0 & 2 \\2 & 0 & 1 \end{bmatrix}$$ Which is the bottom-left matrix. Doing this procedure to the right column gives you a matrix with the values of C.

Sorry for all the math jargon in my answer, if it caused any trouble.

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  • $\begingroup$ Hello! Welcome to the Puzzling Stack Exchange (PSE). Congratulations on your very first answer, here, and it is a great one too — it is thorough; well-explained; and easy to understand. However, since you are new, I advise that you visit the Help Center and, since you have not asked a question as of yet, I highly suggest you go here and here to look at info about questions. Otherwise, keep puzzling! $$\stackrel{\bullet\,\bullet}{\smile}$$ $\endgroup$ – Mr Pie Jul 16 '18 at 22:04

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