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 cases, and therefore the answer is <math>121-2 = \boxed{D}</math> | + | 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 cases, and therefore the answer is <math>121-2 = \boxed{D}</math> |
- Robin's solution | - Robin's solution | ||
Revision as of 14:26, 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 cases, and therefore the answer is 
- Robin's solution
See Also
| 2019 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 13 |
Followed by Problem 15 |
| 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 | |
