# multiplicative inverses in modular arithmetic - breaking up a modulus

Let $p_1$ and $p_2$ be distinct primes. Is it always true that the solution of the simultaneous congruence $$ax \equiv 1 \mod{p_1}$$ $$ax \equiv 1 \mod{p_2}$$ is a solution to $ax \equiv 1 \mod{p_1p_2}?$

I think yes. If $p_1 \mid (aX-1)$ and $p_2 \mid (aX-1)$ then $p_1 p_2=lcm(p_1,p_2) \mid (aX-1)$.