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I'm trying to write a proof but I have been stuck for the past few hours trying to make progress.

I have to prove that

For all integers $n$, $m$, and $p$, if $n| (72m–p)$ and $n| (8m)$, then $n|p$.

I tried substituting $72m-p$ for $x$, and $8m$ for $y$, but I'm not really sure where to go from there.

Bill Dubuque
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Cackling Muse
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  • $72m=9\cdot 8m$. If $n \mid a $ and $n \mid b$ then $n \mid ax+by$ for any integers $x,y$. Note that $p = 9\cdot 8m-(9\cdot 8m-p)$. – copper.hat Mar 26 '21 at 02:36
  • To prove if $,n\mid 8m,$ then $,n\mid 9(8m)-p\Rightarrow n\mid p,$ see the methods in first dupe for proofs via divisibility laws, and see this answer in the second dupe for congruence based proofs and conceptual remarks, – Bill Dubuque Mar 26 '21 at 09:02

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$${\displaystyle 72m-p\equiv 0{\pmod {n}}}$$ $${\displaystyle 9(8m)\equiv p{\pmod {n}}}$$ $${\displaystyle 8m\equiv 0{\pmod {n}}}$$ $${\displaystyle \text{Therefore: }9(0)\equiv p{\pmod {n}}}$$ $${\displaystyle p\equiv 0{\pmod {n}}}$$ $$\rightarrow n|p$$

Bertrand Einstein IV
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  • I dont understand how the mod works or why everything is equal to 0 – Cackling Muse Mar 26 '21 at 13:24
  • @CacklingMuse An equivalence under (mod n) means the LHS and the RHS have equal remainders upon being divided by n. Since n divides $72m-p$ and $8m$ then we can say they are both equal to 0 mod n. Using this we can rearrange to find p=0 (mod n), which means p divided by n yields remainder 0, hence p is divisible by n. – Bertrand Einstein IV Mar 26 '21 at 16:31