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Let d(n) equal the number of positive divisors of the integer n.

Find d(d(p^{p-1})) where p is any prime number.

Example:

$\small{ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline \text{divisors of } n & 1 & 1, 2 & 1, 3 & 1, 2, 4 & 1, 5 & 1, 2, 3, 6 & 1, 7 & 1, 2, 4, 8 & 1, 3, 9 & 1, 2, 5, 10 & 1, 11 & 1, 2, 3, 4, 6, 12 \\ \hline d(n) & 1 & 2 & 2 & 3 & 2 & 4 & 2 & 4 & 3 & 4 & 2 & 6 \\ \hline \end{array} }$

Formula:

d(p) = 2, if p is any prime number.

Formula:

$\begin{array}{|llcll|} \hline \text{if} & n = p_1^{e_1}\cdot p_2^{e_2}\cdot \ldots \cdot p_r^{e_r} \\ \text{then} & d(n) = (e_1+1)(e_2+2)\cdots (e_r+1) \\ \hline \end{array}$

Solution:

$\begin{array}{|rcll|} \hline d(p^{p-1}) &=& (p-1+1) \\ &=& p \\\\ d(p) &=& 2 \\\\ \mathbf{d(d(p^{p-1}))} & \mathbf{=} & \mathbf{2} \\ \hline \end{array}$