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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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Is, according to your definitions, $0 \in N$? –  Johannes Kloos Apr 23 '12 at 15:43
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Depends on whether your math teacher considers $0$ a natural number, but most set theory books do. –  Thomas Andrews Apr 23 '12 at 15:43
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I edited for formatting (you had $<=$ for example. Just tweaks to the format. You can click on the "Edited..." text to get a diff of the edit. –  Thomas Andrews Apr 23 '12 at 15:55
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@JamieB: Because then she would deprive you of the exercise of figuring out for yourself that you were missing 0. Remember, the point of an exercise is usually not to get the solution, but to learn how to solve problems. You don't get any problem solving experience if you're simply told what you did wrong, rather than working it out for yourself. –  Hurkyl Apr 23 '12 at 16:04
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Math teachers may consider 0 a natural number, but number theorists know that it isn't. –  Gerry Myerson Apr 24 '12 at 1:36

1 Answer 1

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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$5$ is odd. Therefore it is odd if you get $5$ as the last digit when you multiply two consecutive digits. –  celtschk Aug 21 '12 at 8:31
    
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 $5 \mid (n+1)$ –  DanZimm Jun 6 '13 at 4:23

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