6. Let $N=69^{5}+5 \times 69^{4}+10 \times 69^{3}+10 \times$ $69^{2}+5 \times 69+1$. Then the number of positive integer divisors of $N$ is ( ) .
(A)3
(B) 216
(C) 69
(D) 5
6. (B).

Using the inverse of the binomial theorem, we get
$$
N=(69+1)^{5}=70^{5}=2^{5} \times 5^{5} \times 7^{5} \text {. }
$$

From this, we can see that the positive integer divisor $a$ of $N$ has the following form:
$$
\begin{array}{l}
a=2^{x} 5^{y} 7^{z}, \\
0 \leqslant x \leqslant 5,0 \leqslant y \leqslant 5,0 \leqslant z \leqslant 5 .
\end{array}
$$
$x, y, z$ each have 6 values to choose from, so the total number of positive integer divisors is $6^{3}=216$.