Difference between revisions of "2020 AMC 10A Problems/Problem 15"
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The prime factorization of <math>12!</math> is <math>2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11</math>. | The prime factorization of <math>12!</math> is <math>2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11</math>. | ||
This yields a total of <math>11 \cdot 6 \cdot 3 \cdot 2 \cdot 2</math> divisors of <math>12!.</math> | This yields a total of <math>11 \cdot 6 \cdot 3 \cdot 2 \cdot 2</math> divisors of <math>12!.</math> | ||
− | In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization. Thus, there are <math>6 \cdot 3 \cdot 2</math> perfect squares. (For <math>2</math>, you can have <math>0</math>, <math>2</math>, <math>4</math>, <math>6</math>, <math>8</math>, or <math>1</math>0 <math>2</math>s, etc | + | In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization. Note that <math>7</math> and <math>11</math> can not be in the prime factorization of a perfect square because there is only one of each in <math>12!.</math> Thus, there are <math>6 \cdot 3 \cdot 2</math> perfect squares. (For <math>2</math>, you can have <math>0</math>, <math>2</math>, <math>4</math>, <math>6</math>, <math>8</math>, or <math>1</math>0 <math>2</math>s, etc.) |
The probability that the divisor chosen is a perfect square is 1/22. m + n = 1 + 22 = 23 <math>\implies \boxed{\textbf{(E) } 23 }</math> | The probability that the divisor chosen is a perfect square is 1/22. m + n = 1 + 22 = 23 <math>\implies \boxed{\textbf{(E) } 23 }</math> | ||
Revision as of 22:08, 31 January 2020
Problem
A positive integer divisor of is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as , where and are relatively prime positive integers. What is ?
Solution
The prime factorization of is . This yields a total of divisors of In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization. Note that and can not be in the prime factorization of a perfect square because there is only one of each in Thus, there are perfect squares. (For , you can have , , , , , or 0 s, etc.) The probability that the divisor chosen is a perfect square is 1/22. m + n = 1 + 22 = 23
See Also
2020 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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All AMC 10 Problems and Solutions |
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