# Divides proof

• Aug 19th 2012, 08:53 PM
anonymouse
Divides proof
I'm struggling with the following question:

let a, b, c, k, l be integers
Prove that if: a|b and a|c then a|(kb + lc)

I'm not looking for the answer outright just a little help.

I realise that:
b = ka
c = la

(kb + lc) = m x a

The approach I've been taking so far is to go: b + c = ka + kl and then try to rearrange (ka + kl) to (kb + lc) since I know that
if a|b and a|c then a|(b + c)

Am I on the right track? If so then I know it's my algebra skills which are the problem.
• Aug 19th 2012, 09:10 PM
richard1234
Re: Divides proof
$k$ and $l$ are arbitrary integers. It doesn't make sense to let $b = ka$ and $c = la$. Plus it confuses readers like me....

You can use different integers, e.g. $b = qa$ and $c = ra$, in which $kb + lc = k(qa) + l(ra) = a(kq + lr)$, clearly, $a$ divides this.

An alternate solution is to use mods, which works just as well (and it doesn't need random variables!).
• Aug 20th 2012, 08:05 PM
anonymouse
Re: Divides proof
Thanks for that Richard, makes much more sense.

We have been studying mods lately so out of interest, how would you approach this using mods?
• Aug 20th 2012, 08:16 PM
richard1234
Re: Divides proof
Basically, $b \equiv 0 (\mod a)$ and $c \equiv 0 (\mod a)$, so for any integers $k$ and $l$,

$bk \equiv 0 (\mod a)$
$cl \equiv 0 (\mod a)$

Therefore the sum $bk + cl$ is also congruent to 0 (mod a).