# Help with a proof - divisor of a linear combination

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• October 17th 2013, 12:30 PM
Stormey
Help with a proof - divisor of a linear combination
Hi guys.

The teacher in our class made some claim and didn't prove it for some reason, and I really want to see the proof for that.

let $n,m,d\in \mathbb{Z}$.
if $d|n$ and $d|m$ then $d|(am+bn)$ for every $a,b\in\mathbb{Z}$.

thanks in advanced.
• October 17th 2013, 12:37 PM
SlipEternal
Re: Help with a proof - divisor of a linear combination
$d|n$ means $n = dx$ for some $x \in \mathbb{Z}$. $d|m$ means $m = dy$ for some $y \in \mathbb{Z}$. So, $am+bn = a(dy) + b(dx) = d(ay+bx)$. Since $ay+bx \in \mathbb{Z}$, $am+bn$ is divisible by $d$.