Math Help - more gcd

1. more gcd

I am stuck on this one too........let m,n be integers such that gcd(m,n)=1. prove that if a equivalent to b mod m and a equivalent to b mod n, then a equivalent to b mod (mn)

thanks

2. Since $a \equiv b \mod m$, we have $m \mid (a-b)$. Similarily $n \mid (a-b)$. Now you probably know that if $s \mid u$, $t \mid u$ and $(s,t)=1$ then $st \mid u$. What do you conclude?

3. ?

Originally Posted by Bruno J.
Since $a \equiv b \mod m$, we have $m \mid (a-b)$. Similarily $n \mid (a-b)$. Now you probably know that if $s \mid u$, $t \mid u$ and $(s,t)=1$ then $st \mid u$. What do you conclude?
do you conclude that a is equivalent to b mod (mn)?

4. Ultimately, yes. But do you understand why?

5. Originally Posted by Bruno J.
Ultimately, yes. But do you understand why?
is it because s and t both divide u therefore mn divides a and b

6. Yes!