Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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...

share|cite|improve this question
Applying division algorithm seems right. – Swapnil Tripathi Jul 5 '14 at 7:56
Look at the last digit of all multiples of $5$:{$5,10,15,20,25,30,....$} or write: $52=5(10)+2$ – user99680 Jul 5 '14 at 7:57
up vote 7 down vote accepted

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.

share|cite|improve this answer

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)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.