$29 \cdot 11$ numbers 1059, 1417, and 2312 each divided by $d$, if the remainder is $r$, where $d$ is an integer greater than 1, then $d-r$ equals
(A) 1 .
(B) 15 .
(C) 179 .
(D) $d-15$.
(E) $d-1$.
(27th American High School Mathematics Examination, 1976)
[Solution] Since $d$ can divide $2312-1417=5 \cdot 179$, and can also divide $1417-1059=2 \cdot 179$, and 179 is a prime number, it is known that $d=179, r=164$, so $d-r=15$. Therefore, the answer is $(B)$.