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$\{n | n \in \mathbb N \text{ and } n\cdot n + n \text{ is a multiple of } 5 \text { and } n \leq 12\}$

I put $\{4,5,9,10\}$ but apparently this is still a proper set not the complete answer?

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    @Gerry: Rule #1 implies a contradiction; therefore Rule #2 is consistent relative to it.2012-10-27

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To make the search easier, I would think of $n*n +n$ as $n * ( n + 1)$, so then you just have to multiply consecutive digits together as see if there is a $5$ as the last digit. This gives the four values you give, along with 0.

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    long with thinking of this as $n(n+1)$ remember that if $k\mid ab$ then either $k \mid a$ or $k \mid b$ so in order to $5 \mid n(n+1)$ either $5\mid n$ or $5 \mid (n+1)$2013-06-06