Difference between revisions of "2011 AMC 12B Problems/Problem 15"
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==Problem 15== | ==Problem 15== | ||
− | How many positive two- | + | How many positive two-digit integers are factors of <math>2^{24}-1</math>? |
<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> | ||
+ | ~ pi_is_3.14 | ||
== Solution == | == Solution == | ||
− | + | Repeating [[difference of squares]]: | |
− | <math>2^{24}-1 = (2^{12} + 1)(2^{6} + 1)(2^{3} + 1)(2^{3} - 1)</math> | + | <math>2^{24}-1=(2^{12}+1)(2^{6}+1)(2^{3}+1)(2^{3}-1)</math> |
− | <math>2^{24}-1 = (2^{12} + 1) | + | <math>2^{24}-1=(2^{12}+1)\cdot65\cdot9\cdot7</math> |
<math>2^{24}-1 = (2^{12} +1) * 5 * 13 * 3^2 * 7</math> | <math>2^{24}-1 = (2^{12} +1) * 5 * 13 * 3^2 * 7</math> | ||
− | + | The sum of cubes formula gives us: | |
− | <math>2^{12} | + | <math>2^{12}+1=(2^4+1)(2^8-2^4+1)</math> |
− | <math>2^{ | + | <math>2^{12}+1 = 17\cdot241</math> |
− | A quick check shows <math> | + | 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. |
− | + | 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>. | |
− | + | 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. | |
− | Multiply <math>7</math> by <math>3, 5 | + | 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. |
− | Multiply <math>5</math> by <math>3</math> or <math>3^2</math> for a two digit factor. | + | 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. |
− | Thus, the total number of factors is <math>3 + 4 + 3 + 2 = \boxed{ | + | Thus, the total number of factors is <math>3+4+3+2=\boxed{\textbf{(D) }12}</math> |
+ | |||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/mgEZOXgIZXs?t=770 | ||
+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/5f4yNbRtDOA | ||
+ | |||
+ | ~savannahsolver | ||
== 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}} | ||
+ | |||
+ | [[Category:Introductory Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 19:54, 29 May 2023
Contents
Problem 15
How many positive two-digit integers are factors of ?
~ pi_is_3.14
Solution
Repeating difference of squares:
The sum of cubes formula gives us:
A quick check shows is prime. Thus, the only factors to be concerned about are , since multiplying by will make any factor too large.
Multiplying by or will give a two-digit factor; itself will also work. The next smallest factor, , gives a three-digit number. Thus, there are factors that are multiples of .
Multiplying by , , or will also give a two-digit factor, as well as itself. Higher numbers will not work, giving additional factors.
Multiply by , , or for a two-digit factor. There are no more factors to check, as all factors which include are already counted. Thus, there are an additional factors.
Multiply by or for a two-digit factor. All higher factors have been counted already, so there are more factors.
Thus, the total number of factors is
Video Solution by OmegaLearn
https://youtu.be/mgEZOXgIZXs?t=770
Video Solution by WhyMath
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.