$29 \cdot 13$ has more than one positive integer greater than 1, which when divided by any integer $k(2 \leqslant$ $k \leqslant 11)$ leaves a remainder of 1. The difference between the two smallest such integers is
(A) 2310.
(B) 2311.
(C) 27720.
(D) 27721.
(E) None of these.
(29th American High School Mathematics Examination, 1978)
[Solution] If two positive integers, when divided by any integer from 2 to 11, leave a remainder of 1, then the integers are of the form $m t+1$, where $t$ is a non-negative integer, and $m=2^{3} \cdot 3^{2} \cdot 5 \cdot 7$. $11=27720$ is the least common multiple of $2,3, \cdots, 11$.
Therefore, the difference between the two smallest integers with the desired property is 27720.
Hence, the answer is $(C)$.