$$
\begin{aligned}
\text { 23. Let } N=69^{5}+5 \cdot 69^{4}+10 \cdot 69^{3}+10 \cdot 69^{2} \\
+5 \cdot 69+1
\end{aligned}
$$

How many positive integers are divisors of $N$?
(A) 3; (B) 5 ; (C) 69; () 125; (D) 216 .

$\frac{23}{\mathrm{E}}$
23. By the binomial theorem, $N=(69+1)^{5}$ $=(2,5,7)^{5}=2^{5}, 5^{5}, 7^{5},$ the number of factors of $N$ is $(5+1)(5+1)(5+1)=6^{3}=216$.