$30 \cdot 40$ Let $N$ be an integer, the smallest positive integer $x$ such that $1260 x=N^{3}$ is
(A) 1050.
(B) 1260.
(C) $1260^{2}$.
(D) 7350.
(E) 44100.
(14th American High School Mathematics Examination, 1963)
[Solution] From $1260 x=2^{2} \cdot 3^{2} \cdot 5 \cdot 7 \cdot x$,
thus, if $1260 x=N^{3}$, it is only necessary for the exponents of all distinct prime factors to be 3 for it to be a cube number.
Therefore, the smallest $x$ is $\quad x=2 \cdot 3 \cdot 5^{2} \cdot 7^{2}=7350$.
Hence, the answer is $(D)$.