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If I suppose that $5$ divides $52$, then there would exist an $ s \in \mathbb Z $ such that $ 5s = 52 $. There is no such s, because $5(10) = 50$, and $5(11) = 55$. I'm not convinced with this proof, because I beleive I'm missing an argument, ad I´m am not sure what is it.

Another way I was thinkng was to say that, because of the division algorithm, $5(10) + 2 = 52$, which would prove that $2$ is different from $0$, ad therefore there is no $ s \in \mathbb Z $ such that $ 5s = 52 $. I am still not convinced...

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    $\begingroup$ Applying division algorithm seems right. $\endgroup$ – Swapnil Tripathi Jul 5 '14 at 7:56
  • $\begingroup$ Look at the last digit of all multiples of $5$:{$5,10,15,20,25,30,....$} or write: $52=5(10)+2$ $\endgroup$ – user99680 Jul 5 '14 at 7:57
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When a nonzero integer $y$ divides an integer $x$, it is a theorem that the division algorithm produces a unique remainder $r$ and quotient $q$ such that $x = qy + r$ for $0 \leq r < y$. Can you prove it? Hint: proceed by contradiction.

Dividing $5$ into $52$, we get $52 = 10 \cdot 5 + 2$.

Note that $y|x$ if and only if $r = 0$, which it does not in this case.

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You could maybe prove the following: if $s>11$, then $5s>55$. Further, $52$ is not more than $55$. So, if $5s=52$, then $s\leq 11$ and you can check the $11$ possibilities by hand. (However, you need to have order on $\mathbb{N}$ for this argument)

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