Answer
See explanation
Work Step by Step
**Proof:**
Let \( m \), \( d \), and \( k \) be integers with \( d > 0 \). By the division algorithm, we can write
\[
m = qd + r,
\]
where \( q \) is an integer and \( r \) is the remainder, satisfying
\[
0 \le r < d.
\]
By definition, \( m \mod d = r \).
Now, consider
\[
m + d \cdot k = qd + r + d \cdot k = (q + k)d + r.
\]
When we take the modulo \( d \) of this sum, the term \((q+k)d\) is divisible by \( d \) and thus contributes nothing to the remainder. Therefore,
\[
(m + d \cdot k) \mod d = r.
\]
Since \( m \mod d = r \) as well, we have
\[
(m + d \cdot k) \mod d = m \mod d.
\]
This completes the proof.
