$29 \cdot 6$ a number when divided by 10 leaves a remainder of 9; when divided by 9 leaves a remainder of 8; when divided by 8 leaves a remainder of 7; and so on until when divided by 2, it leaves a remainder of 1. The number could be
(A) 59.
(B) 419.
(C) 1259.
(D) 2519.
(E) None of the above.
(2nd American High School Mathematics Examination, 1951)
[Solution] Let this number be $N$, then $N+1$ is a multiple of $10,9,8, \cdots, 2$, so we can take their least common multiple:
$$
[10,9,8, \cdots, 2]=2^{3} \cdot 5 \cdot 3^{2} \cdot 7=2520,
$$

thus $N=2520-1=2519$.
Therefore, the answer is $(D)$.