Difference between revisions of "2020 AMC 10A Problems/Problem 15"
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<math>\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 23</math> | <math>\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 23</math> | ||
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| + | ==Solution== | ||
| + | The prime factorization of 12! is <math>2^{10}</math> * <math>3^5</math> * <math>5^2</math> * <math>7</math> * <math>11</math>. | ||
| + | This yields a total of <math>11</math> * <math>6</math> * <math>3</math> * <math>2</math> * <math>2</math> divisors of 12!. | ||
| + | In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization. Thus, there are 5 * 3 * 2 perfect squares. (For 2, you can have 0, 2, 4, 6, 8, or 10 2s, etc. Note that 7 and 11 can not be in the prime factorization of a perfect square because there is only one of each in 12!.) | ||
| + | The probability that the divisor chosen is a perfect square is 1/22. m + n = 23. | ||
==See Also== | ==See Also== | ||
Revision as of 23:01, 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 12! is 
* 
* 
* 
* 
.
This yields a total of * 
* 
* 
* divisors of 12!.
In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization. Thus, there are 5 * 3 * 2 perfect squares. (For 2, you can have 0, 2, 4, 6, 8, or 10 2s, etc. Note that 7 and 11 can not be in the prime factorization of a perfect square because there is only one of each in 12!.)
The probability that the divisor chosen is a perfect square is 1/22. m + n = 23.
See Also
| 2020 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
