Congruence and modular arithmetics

Question

I started to study number theory and got stuck in the following exercise:

Let $a,b\in \mathbb{Z}$. Show that $ab\equiv lcm(a,b)(\mbox{mod gcd}(a,b))$.

I have no idea how to show that. Any help?

Thanks!

Answer 1

Hint.

$a,b$ are both divisible by $gcd(a,b)$ so $ab\equiv0 \pmod {gcd(a,b)}$.
the $lcm$ is $kab$ where $k$ is a real number for witch $kab \geq a$ if $a<b$ and $gcd \leq a$.

Answer 2

Hint: Let $d=gcd(a,b)$ and $m=lcm(a,b)$. Write $a=a'd$ and $b=b'd$. Write $m$ in terms of $a',b',d$.