2011 AMC 12B Problems/Problem 15

Problem 15

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

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

Solution

From repeated application of difference of squares:

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

$2^{24}-1 = (2^{12} + 1) * 65 * 9 * 7$

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

Aplying sum of cubes:

$2^{24}-1 = (2^4 - 1)(2^8 + 2^4 + 1) * 3^2 * 5 * 7 * 13$

$2^{24}-1 = 17 * 241 * 3^2 * 5 * 7 * 13$

A quick check shows $241$ is prime. Thus, the only factors to be concerned about are $3^2 * 5 * 7 * 13 * 17$ , since multiplying by $241$ will make any factor too large.

Multiply $17$ 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 which are multiples of $17$ .

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

Multiply $7$ by $3, 5,$ or $3^2$ for a two digit factor. There are no mare 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 $3 + 4 + 3 + 2 = \boxed{12 \textbf{ (D)}}$

2011 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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2011 AMC 12B Problems/Problem 15

Problem 15

Solution

See also