I have to do it without using Fundamental Theorem of Arithmetic. Can someone check my work?
Prove if $2\mid x$ and $5\mid x$, then $10\mid x$.
Let $x \in \mathbb{Z}$. Suppose $2\mid x$ and $5\mid x$. Then, by definition of divisibility, there exists integers $m,n$ such that $2m = x = 5n$. Therefore $5\mid 2m$. Since the $\gcd(2,5) = 1$ it follows that $5\mid m$. Therefore there exists an integer $k \in \mathbb{Z}$ such that $5k = m$.
Prove if $10\mid x$, then $2\mid x$ and $5\mid x$.
Suppose $10\mid x$, then there exists an integer $a \in \mathbb{Z}$ such that $10a = x$. From here it follows that $2 \cdot 5 \cdot a = x$.
I really don't feel like this encompasses the entire proof. Should I be using Euclid's Lemma instead?