a = b (mod m) and a = b (mod n) iff a = b (mod lcm(m,n))?

Question

Basically I'm wondering if the converse of this is true:

If $a \equiv b (mod m)$ and $a \equiv b (mod n)$ then $a \equiv b (mod lcm(m, n))$

Answer 1

$a=b+{mnk\over gcd(m,n)}$ for some integer $k$, hence $a=b+m{nk\over gcd(m,n)}=b+n{mk\over gcd(m,n)}$ and hence the converse is true.