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If you have a cube, say, $3 \times 3 \times 3$, you know that mathematically it's made of $27$ ($3*3*3$) smaller cubes. Now, just remove one full line of smaller cubes (in this case, 3 smaller cubes).

Can you prove that the remaining number of smaller cubes is always a multiple of 6?

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  • $\begingroup$ 24 = 6*4. What is the problem? $\endgroup$
    – klm123
    Aug 31, 2014 at 8:40
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    $\begingroup$ @klm123 You mean the question has an ambiguity in that it doesn't say "by always I meant all cubes, not just 3x3x3"? In that case, I'd say that was a bit far-fetched. $\endgroup$
    – Mr Lister
    Aug 31, 2014 at 18:57
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    $\begingroup$ I see. "Always" means: for any integer value of the symbol "3". $\endgroup$
    – Florian F
    Sep 3, 2014 at 14:26
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    $\begingroup$ Why would you see (edit) and to not accept my answer? I see it as completely answering your question, if not, please specify what else would you like to know. It is important on beta site for answers to be accepted if they are adequate. $\endgroup$
    – klm123
    Sep 29, 2014 at 11:44

1 Answer 1

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We will have $N = x^3 - x = x(x-1)(x+1)$ small cubes left.

One of the x and (x-1) is divisible by 2. One of the (x-1), x and (x+1) is divisible by 3. Therefore N is divisible by 6.

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