Vtyjkyuk

I would like to prove if $a mid n$ and $b mid n$ then $a cdot b mid n$ for $forall n ge a cdot b$ where $a, b, n in mathbbZ$

I'm stuck.

$n = a cdot k_1$

$n = b cdot k_2$

$therefore a cdot k_1 = b cdot k_2$

EDIT: so for fizzbuzz it wouldn't make sense to check to see if a number is divisible by 15 to see if it's divisible by both 3 and 5?

You are possibly thinking of the following: if $amid n$ and $bmid n$ and $a,b$ are relatively prime (have no common factor except 1), then $abmid n$.

Proof. We have $n=ak$ and $n=bl$ for some integers $k,l$. Therefore $bmid ak$; since $a,b$ are relatively prime this implies $bmid k$, so $k=bm$, so $n=abm$; therefore $abmid n$.

Re: edit, yes this would make sense because $3$ and $5$ are relatively prime.

This is false. For example, 3 | 30 and 6 | 30, but their product, 18, does not divide 30 even though $3 times 6 < 30$.

Updated: The edited version is still not true. At least two counterexamples in the comments below.

Prior counterexample: This is not true. $12|36$ and $9|36$, but $12cdot9 = 108 not | 36$.

Hint $, a,bmid niff rm lcm(a,b)mid n!!!overset large times, (a,b)iff abmid n(a,b),, $ which is not equivalent to $,abmid n $