Proving that if $3| x$ and $5|x$, then $15|x$.

Question

Prove if $3| x$ and $5|x$, then $15|x$.

I'm thinking the answer will have something to do with either the fact that both $3$ and $5$ are prime, or that $15$ is the lowest common multiple of $3$ and $5$, but I'm not sure how to use those facts in the proof.

Answer 1

Write $x=3k=5j$. Then $3|5j$. Since $3\not\mid 5$ we have that $3|j$ since $3$ is prime. Hence $x=5\cdot 3\cdot j'=15j'$. Hence $15|x$.

Answer 2

From the fact that $y$ divides $x$ and $z$ divides $x$ you cannot infer that $yz$ divides $x$: $3$ divides $6$ and $3$ divides $6$, but $9$ does not divide $x$. But you're close: it's the fact that $3$ and $5$ are co-prime that makes their product divide $x$.

Answer 3

Just use the danged definitions.

$3|x$ means that there is some $k$ so that $x = 3k$ and $5|x$ means that $x=5j$ for some $j$. So $3k = 5j$ so $k = \frac{5j}3$ is an integer. But $3$ and $5$ have no factors in common, so $3|j$. so $j = 3m$ for some $m$. So $x = 5j = 5*3*m = 15m$. So $15|m$.

Answer 4

If $3$ and $5$ divide $x$, the latter is a common multiple of $3$ and $5$, hence a multiple of $\;\operatorname{lcm}(3,5)$. Now, since $3$ and $5$ are coprime, $$\operatorname{lcm}(3,5)=3\cdot 5.$$