No calculators allowed!

Let $f(x) = \left((1+x)^{101}-1\right)^5 = \binom 50 (1+x)^{505} - \binom 51 (1+x)^{404} + \binom 52 (1+x)^{303} - \binom 53 (1+x)^{202} + \binom 54 (1+x)^{101} - 1$ .

Note that $N$ is the coefficient of $x^5$ of $f(x)$ or $N = \dfrac {f^{(5)} (0)}{5!}$ , where $f^{(n)}$ is the $n$ th derivative of $f(x)$ .

Since $f^{(5)} (x) = 5!101^5 \implies N = 101^5$ . Noting that 101 is a prime, the number of divisors of $N$ is $5+1=\boxed 6$ .

$N$ is equal to the coefficient of $x^{5}$ in

$\binom{5}{0}(1+x)^{505}-\binom{5}{1}(1+x)^{404}+\binom{5}{2}(1+x)^{303}-\binom{5}{3}(1+x)^{202}+\binom{5}{4}(1+x)^{101}-\binom{5}{5}$

$= ((1+x)^{101}-1)^{5}$

$=x^{5}(101+\binom{101}{2}x+\binom{101}{3}x^{2}+\dots+\binom{101}{100}x^{99}+x^{100})^{5}$

Clearly, the coefficient of $x^{5}$ in this simplified expression is given by just $101^{5}$ . Thus, the number of positive integral divisors of $N$ is $\boxed{6}$ .