## Discrete Mathematics with Applications 4th Edition

$4$
If $b\ mod\ 12=5$, then $b=12q+5$ for some integer $q$. Multiplying both sides by $8$, we get $8b=8(12q+5)=12(8q)+40$$=12(8q)+36+4=12(8q)+12(3)+4$$=12(8q+3)+4$. Since $0\leq4\lt12$, and since $8q+3$ is an integer, it must be that $4=8b\ mod\ 12$.