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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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| − | == 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 == | | == See also == |
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