+0  
 
0
43
1
avatar+143 

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.

waffles  Aug 23, 2017
Sort: 

1+0 Answers

 #1
avatar+18564 
0

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:

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

 

 

laugh

heureka  Aug 23, 2017

5 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details