$32 \cdot 46$ If we set
$$
N=69^{5}+5 \cdot 69^{4}+10 \cdot 69^{3}+10 \cdot 69^{2}+5 \cdot 69+1 .
$$

The number of positive integer divisors of $N$ is
(A) 3 .
(B) 5 .
(C) 69 .
(D) 125 .
(E) 216
(37th American High School Mathematics Examination, 1986)
[Solution] Notice that the coefficients of the expansion of $(a+b)^{5}$ are $1,5,10,10,5,1$, then
$$
N=(69+1)^{5}=70^{5}=2^{5} \cdot 5^{5} \cdot 7^{5} \text {. }
$$

Thus, the number of positive integer divisors of $N$ is
$$
(5+1)(5+1)(5+1)=6^{3}=216 \text { (divisors). }
$$

Therefore, the answer is $(E)$.