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

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==Problem 15==
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How many positive two-digit integers are factors of <math>2^{24}-1</math>?
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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>
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~ pi_is_3.14
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== Solution ==
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Repeating [[difference of squares]]:
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<math>2^{24}-1=(2^{12}+1)(2^{6}+1)(2^{3}+1)(2^{3}-1)</math>
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<math>2^{24}-1=(2^{12}+1)\cdot65\cdot9\cdot7</math>
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<math>2^{24}-1 = (2^{12} +1) * 5 * 13 * 3^2 * 7</math>
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The sum of cubes formula gives us:
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<math>2^{12}+1=(2^4+1)(2^8-2^4+1)</math>
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<math>2^{12}+1 = 17\cdot241</math>
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A quick check shows <math>241</math> is prime. Thus, the only factors to be concerned about are <math>3^2\cdot5\cdot7\cdot13\cdot17</math>, since multiplying by <math>241</math> will make any factor too large.
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Multiplying <math>17</math> by <math>3</math> or <math>5</math> will give a two-digit factor; <math>17</math> itself will also work. The next smallest factor, <math>7</math>, gives a three-digit number.  Thus, there are <math>3</math> factors that are multiples of <math>17</math>.
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Multiplying <math>13</math> by <math>3</math>, <math>5</math>, or <math>7</math> will also give a two-digit factor, as well as <math>13</math> itself. Higher numbers will not work, giving <math>4</math> additional factors.
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Multiply <math>7</math> by <math>3</math>, <math>5</math>, or  <math>3^2</math> for a two-digit factor. There are no more factors to check, as all factors which include <math>13</math> are already counted. Thus, there are an additional <math>3</math> factors.
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Multiply <math>5</math> by <math>3</math> or <math>3^2</math> for a two-digit factor. All higher factors have been counted already, so there are <math>2</math> more factors.
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Thus, the total number of factors is <math>3+4+3+2=\boxed{\textbf{(D) }12}</math>
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== Video Solution by OmegaLearn ==
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https://youtu.be/mgEZOXgIZXs?t=770
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==Video Solution by WhyMath==
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https://youtu.be/5f4yNbRtDOA
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~savannahsolver
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== See also ==
 
== See also ==
 
{{AMC12 box|year=2011|ab=B|num-b=14|num-a=16}}
 
{{AMC12 box|year=2011|ab=B|num-b=14|num-a=16}}

Latest revision as of 19:54, 29 May 2023

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


Video Solution by OmegaLearn

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

Video Solution by WhyMath

https://youtu.be/5f4yNbRtDOA

~savannahsolver

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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