Suppose that $d,\, m > 1$, and that $d|m$. Let $a,\,b \in\mathbb{ Z}$, prove that if $a \cong b (mod\,m)$ then $a \cong b (mod\,d)$.
$a\equiv\bar a\!\pmod{\!kn}\Rightarrow a\equiv\bar a\!\pmod{\! n};\ $ $(a\bmod kn)\bmod n=a\bmod n.\ $ Congruences persist mod factors of the modulus