$5 \cdot 1$ If $r$ is the remainder when 1059, 1417, and 2312 are divided by $d$, where $d$ is an integer greater than 1, find the value of $d-r$.
(29th International Mathematical Olympiad Shortlist, 1988)
［Solution］From the given $1059 \equiv 1417 \equiv 2312(\bmod d)$.
Then
$$
\begin{array}{l}
d \mid 2312-1417=895, \\
d \mid 1417-1059=358 .
\end{array}
$$

And
$$
(895,358)=179 \text {, }
$$

So we have $d \mid 179$.
Since 179 is a prime number, $d>1$, thus
$$
d=179 .
$$

From
$$
1059=179 \cdot 5+164,
$$

Therefore
$$
\begin{aligned}
r & =164 . \\
d-r=179-164 & =15 .
\end{aligned}
$$