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| − | ==Problem==
| + | #REDIRECT[[2019_AMC_10B_Problems/Problem_19]] |
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| − | Let <math>S</math> be the set of all positive integer divisors of <math>100,000.</math> How many numbers are the product of two distinct elements of <math>S?</math>
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| − | ==Solution==
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| − | 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
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| − | ==See Also==
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| − | {{AMC12 box|year=2019|ab=B|num-b=13|num-a=15}}
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