# 2011 AMC 12B Problems/Problem 15

## Problem 15

How many positive two-digit integers are factors of $2^{24}-1$?

$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 14$

~ pi_is_3.14

## Solution

Repeating difference of squares:

$2^{24}-1=(2^{12}+1)(2^{6}+1)(2^{3}+1)(2^{3}-1)$

$2^{24}-1=(2^{12}+1)\cdot65\cdot9\cdot7$

$2^{24}-1 = (2^{12} +1) * 5 * 13 * 3^2 * 7$

The sum of cubes formula gives us:

$2^{12}+1=(2^4+1)(2^8-2^4+1)$

$2^{12}+1 = 17\cdot241$

A quick check shows $241$ is prime. Thus, the only factors to be concerned about are $3^2\cdot5\cdot7\cdot13\cdot17$, since multiplying by $241$ will make any factor too large.

Multiplying $17$ by $3$ or $5$ will give a two-digit factor; $17$ itself will also work. The next smallest factor, $7$, gives a three-digit number. Thus, there are $3$ factors that are multiples of $17$.

Multiplying $13$ by $3$, $5$, or $7$ will also give a two-digit factor, as well as $13$ itself. Higher numbers will not work, giving $4$ additional factors.

Multiply $7$ by $3$, $5$, or $3^2$ for a two-digit factor. There are no more factors to check, as all factors which include $13$ are already counted. Thus, there are an additional $3$ factors.

Multiply $5$ by $3$ or $3^2$ for a two-digit factor. All higher factors have been counted already, so there are $2$ more factors.

Thus, the total number of factors is $3+4+3+2=\boxed{\textbf{(D) }12}$

~savannahsolver