Difference between revisions of "2011 AMC 12B Problems/Problem 15"
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| + | ==Problem 15== | ||
| + | How many positive two-digits 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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| + | == Solution == | ||
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| + | From repeated application of 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) * 65 * 9 * 7</math> | ||
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| + | <math>2^{24}-1 = (2^{12} +1) * 5 * 13 * 3^2 * 7</math> | ||
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| + | Applying sum of cubes: | ||
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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 * 241</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 * 5 * 7 * 13 * 17</math>, since multiplying by <math>241</math> will make any factor too large. | ||
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| + | Multiply <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 which are multiples of <math>17</math>. | ||
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| + | Multiply <math>13</math> by <math>3, 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 an additional <math>4</math> factors. | ||
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| + | Multiply <math>7</math> by <math>3, 5, </math> or <math> 3^2</math> for a two digit factor. There are no mare 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{12 \textbf{ (D)}}</math> | ||
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| + | == See also == | ||
| + | {{AMC12 box|year=2011|ab=B|num-b=14|num-a=16}} | ||
| + | |||
| + | [[Category:Introductory Number Theory Problems]] | ||
| + | {{MAA Notice}} | ||
Revision as of 17:26, 5 February 2018
Problem 15
How many positive two-digits integers are factors of 
?
Solution
From repeated application of difference of squares:
Applying sum of cubes:
A quick check shows 
is prime. Thus, the only factors to be concerned about are 
, since multiplying by will make any factor too large.
Multiply 
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 which are multiples of .
Multiply 
by 
or will also give a two digit factor, as well as itself. Higher numbers will not work, giving an additional 
factors.
Multiply by 
or 
for a two digit factor. There are no mare 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 
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. 
