\begin{problem}

This problem asks you to prove some simple facts about divisibility
and congruence.

\bparts

\ppart Prove: If $a \mid b$, then $a \mid b c$ for all $c$.

\solution[\vspace{2.5in}]{If $a \mid b$, then there exists a $k$ such
that $a k = b$.  Multiplying both sides by $c$ gives $a (k c) = b c$,
which implies that $a \mid b c$.}

\ppart Prove: If $a \mid b$ and $a \mid c$, then $a \mid s b + t c$
for all $s, t$.

\solution[\vspace{2.5in}]{
Suppose that $a \mid b$ and $a \mid c$.  Then there exist
integers $k_1$ and $k_2$ such that $a k_1 = b$ and $a k_2 = c$.  Thus,
$s b + t c = s (a k_1) + t (a k_2) = a (s k_1 + t k_2)$, which implies
that $a \mid s b + t c$.
}

\ppart Prove that congruence is transitive: If $a \equiv b \pmod{n}$
and $b \equiv c \pmod{n}$, then $a \equiv c \pmod{n}$.

\solution[\newpage]{If $a \equiv b \pmod{n}$, then $n \mid (a -
b)$.  If $b \equiv c \pmod{n}$, then $n \mid (b - c)$.  Therefore, $n$
divides $(a - b) + (b - c) = a - c$, which means that $a \equiv c
\pmod{n}$.}

\eparts

\end{problem}