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Proving $d\ell= ad+b$ for integers that $d$ must divide $b$.

If $d\ell$ is a product we can write it as $\underbrace{d+d+d+\ldots}_{\ell \text{ times}}$ and $ad$ as a sum of $\underbrace{d+d+d+\ldots}_{a \text{ times}}$ which means $b$ must be a sum of $d$'s, but how can I prove that to be true?

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    Your question was put on hold, the message above (and possibly comments) should give an explanation why. (In particular, [this link](http://meta.math.stackexchange.com/a/9960) might be useful.) You might try to edit your question to address these issues. Note that the next edit puts your post in the review queue, where users can vote whether to reopen it or leave it closed. (Therefore it would be good to avoid minor edits and improve your question as much as possible with the next edit.) – 2017-01-26

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Move $ad$ to the LHS and factorize $d$ out of both terms on the LHS. Since $d$ divides LHS then it must divide $b$ the only term left on the RHS as well.

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    im an idiot. will accept once it lets me. thanks. – 2017-01-24