10. (10 points) If 2017, 1029, and 725 leave a remainder of $r$ when divided by $d$, then the maximum value of $d-r$ is $\qquad$
【Analysis】According to the problem, $2017-r, 1029-r, 725-r$, can all be divided by $d$, then $(2017-r)-$ $(1029-r),(2017-r)-(725-r),(1029-r)-(725-r)$, these three numbers can also be divided by $d$, i.e., 988, 1292, 304 can all be divided by $d$. It is not difficult to conclude that the greatest common divisor of the three numbers is 76, i.e., the value of $d$ could be: $76,38,19,4,2,1$ (the remainder when divided by 1 can be considered as 0); then, divide 725 by the possible values of $d$, find the value of $d-r$, and select the maximum value of $d-r$.

【Solution】According to the problem, $2017-r, 1029-r, 725-r$, can all be divided by $d$, then $(2017-r)-(1029-r),(2017-r)-(725-r),(1029-r)-(725-r)$, these three numbers can also be divided by $d$, i.e., $988,1292,304$ can all be divided by $d$,
$$
\begin{array}{l}
988=2 \times 2 \times 19 \times 13 \\
1292=2 \times 2 \times 19 \times 17 \\
304=2 \times 2 \times 2 \times 2 \times 19
\end{array}
$$

So the greatest common divisor of the three numbers is: $2 \times 2 \times 19=76$, $d$ is a factor of 76, i.e., the value of $d$ could be: $76,38,19,4,2,1$ (the remainder when divided by 1 can be considered as 0). When $d=76$, at this time: $725 \div 76=9 \cdots 41$, i.e., $r=41$, i.e., at this time $d-r=76-41=35$; when $d=38$, at this time: $725 \div 38=19 \cdots 3$, i.e., $r=3$, i.e., at this time $d-r=38-3=35$; when $d=19$, at this time: $725 \div 19=38 \cdots 3$, i.e., $r=3$, i.e., at this time $d-r=19-3=16$; when $d=4$, at this time: $725 \div 4=182 \cdots 1$, i.e., $r=1$, i.e., at this time $d-r=4-1=3$; when $d=2$, at this time: $725 \div 2=362 \cdots 1$, i.e., $r=1$, i.e., at this time $d-r=2-1=1$; when $d=1$, at this time: $725 \div 1=725$, i.e., $r=0$, i.e., at this time $d-r=1-0=1$; then, the maximum value of $d-r$ is 35. Therefore, the answer is: 35.