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Difference between revisions of "2011 AMC 12B Problems/Problem 15"

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<math>\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 14</math>
 
<math>\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 14</math>
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== Video Solution ==
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https://youtu.be/mgEZOXgIZXs?t=770
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~ pi_is_3.14
  
 
== Solution ==
 
== Solution ==

Revision as of 22:23, 17 January 2021

Problem 15

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

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

Video Solution

https://youtu.be/mgEZOXgIZXs?t=770

~ pi_is_3.14

Solution

From repeated application of difference of squares:

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

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

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

Applying sum of cubes:

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

$2^{12}+1 = 17 * 241$


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 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 $3 + 4 + 3 + 2 = \boxed{12 \textbf{ (D)}}$

See also

2011 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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