If $c | m_i$ Prove that $c| \sum_{i=1}^r u_i m_i$ where $m_i,u_i,c \in \mathbb{Z}$ and $i=1,2,..,r$

Question

Let $m_1,m_1,...,m_r,u_1,u_2,...,u_r,c \in \mathbb{Z}$, such that $c | m_i$ for each $i \in \lbrace 1,2,...,r \rbrace$.

Prove that $c| \sum_{i=1}^r u_i m_i$

My attempt:

We have $c | m_i$ for each $i \in \lbrace 1,2,...,r \rbrace$, then

$c|u_i m_i$ for each $i \in \lbrace 1,2,...,r \rbrace$, thus

$\exists k_i\in \mathbb{Z}$ such that $u_i m_i=k_i c$, $\forall i$

then

$\sum_{i=1}^r u_i m_i=\sum_{i=1}^r k_i c=c \sum_{i=1}^r k_i$

and since $\sum_{i=1}^r k_i \in \mathbb{Z}$ we got that

$c|\sum_{i=1}^r u_i m_i$

Is that true please ?, thanks.

Answer 1

Your proof is entirely correct. You might also use the fact that $c\mid m_i$ to deduce that $m_i=n_ic$ and hence that $u_im_i=u_in_ic$, to reach the same conclusion.