Difference between revisions of "2019 AMC 12B Problems/Problem 14"
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==Solution== | ==Solution== | ||
− | First, find the prime factorization of <math>100,000</math>. It is <math>2^5 \cdot 5^5</math>. Thus, any factor will have the pattern <math>2^m \cdot 5^n</math>, where <math>m, n < 5</math>. Multiplying this by another factor with the same pattern <math>2^k \cdot 5^l</math> gets us <math>2^m+k \cdot 5^n+l</math>. It initially seems like we have <math>11</math> options for the power of <math>2</math> and <math>11</math> options for the power of <math>5</math>, giving us a total of <math>121</math> choices. However, note that the factors must be distinct. If they are distinct, we cannot have <math>1</math> (as it is only formed by <math>1 \cdot 1</math>), or <math>2^{10} \cdot 5^{10}</math> (as it is only formed by <math>100,000 \cdot 100,000</math>). These are the only two cases where the distinction rule forces us to eliminate case, and therefore the answer is <math>121-2</math> D | + | First, find the prime factorization of <math>100,000</math>. It is <math>2^5 \cdot 5^5</math>. Thus, any factor will have the pattern <math>2^m \cdot 5^n</math>, where <math>m, n < 5</math>. Multiplying this by another factor with the same pattern <math>2^k \cdot 5^l</math> gets us <math>2^m+k \cdot 5^n+l</math>. It initially seems like we have <math>11</math> options for the power of <math>2</math> and <math>11</math> options for the power of <math>5</math>, giving us a total of <math>121</math> choices. However, note that the factors must be distinct. If they are distinct, we cannot have <math>1</math> (as it is only formed by <math>1 \cdot 1</math>), or <math>2^{10} \cdot 5^{10}</math> (as it is only formed by <math>100,000 \cdot 100,000</math>). These are the only two cases where the distinction rule forces us to eliminate case, and therefore the answer is <math>121-2</math> \boxed{D} |
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+ | Robin's Solution | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2019|ab=B|num-b=13|num-a=15}} | {{AMC12 box|year=2019|ab=B|num-b=13|num-a=15}} |
Revision as of 13:08, 14 February 2019
Problem
Let be the set of all positive integer divisors of How many numbers are the product of two distinct elements of
Solution
First, find the prime factorization of . It is . Thus, any factor will have the pattern , where . Multiplying this by another factor with the same pattern gets us . It initially seems like we have options for the power of and options for the power of , giving us a total of choices. However, note that the factors must be distinct. If they are distinct, we cannot have (as it is only formed by ), or (as it is only formed by ). These are the only two cases where the distinction rule forces us to eliminate case, and therefore the answer is \boxed{D}
Robin's Solution
See Also
2019 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
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All AMC 12 Problems and Solutions |