Solution f7ba22ed-207d-4522-8b70-1703a98d4ccf

Points: 1 pt

Since $a\mid b_i$, the definition of divisibility gives
\[
  b_i=ak_i \qquad \text{for some } k_i\in\mathbb{Z}.
\]
Multiplying by $c_i$, we obtain
\[
  b_ic_i=a(k_ic_i).
\]
Since $k_ic_i$ is an integer, we have $a\mid b_ic_i$ for every
$i=1,2,\ldots,n$. By additivity of divisibility, $a$ therefore divides
the sum of these terms.

Final answer:
a\mid\sum_{i=1}^{n} b_ic_i