Proving nice divisibility

Question

Let $n=10x+y$ where $n$, $x$ and $y$ are positive integers. Prove that $n$ is divisible by $13$ iff $x+4y$ is divisible by $13$.

I let $n=13k$, thereafter mutliplied $x+4y$ by $10$, to get $10x+40y$ and $n=10x+y$. Subtract the two to get $39y$ which is clearly divisible by $13$, but I don't know how to structure this proof.

@peterwhy I realised how much of a no-brainer it was after working out. — Daiyuan Han, Mar 08 '24 at 04:01

score 1 · Answer 1 · edited Mar 08 '24 at 02:28

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Assume $10x+y=13M$ and $y=13M-10x$. Then: $$x+4y=x+4(13M-10x) =52M-39x=13P.$$ Assume $x+4y=13N x=13N-4y$ Then: $$10x+y=10(13N-4y)+y =130N-39y=13Q.$$

edited Mar 08 '24 at 02:28

Alex Pozo

1,290

answered Mar 08 '24 at 02:05

Daiyuan Han

5
4

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Please use Mathjax to format your answer. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – sudeep5221 Mar 08 '24 at 02:15
no thanks @sudeep5221 immanswering my own qn. – Daiyuan Han Mar 08 '24 at 02:16

score 0 · Answer 2 · answered Mar 08 '24 at 02:23

$n=10x+y. 13|n \iff 13|(x+4y)?$

$(40x+4y)-(x+4y)=39x$

$(x+4y)+39x=4(10x+y)$

If a prime divides two numbers, it divides their sum and their difference.

IF $13|(x+4y)$ and $13|39x$, so $13|4(10x+y)$. Since $13\not|4 \implies 13|10x+y$

By similar reasoning, one can prove the other direction .

Proving nice divisibility

2 Answers2