Difference between revisions of "2019 AMC 10A Problems/Problem 25"
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<cmath>\frac{(n^2)!}{(n!)^{n+1}}=\frac{n^2}{n!}\cdot\frac{(n^2-1)!}{(n!)^n}</cmath> | <cmath>\frac{(n^2)!}{(n!)^{n+1}}=\frac{n^2}{n!}\cdot\frac{(n^2-1)!}{(n!)^n}</cmath> | ||
| − | is not an integer if <math>n^2 \nmid n!</math>. This happens only for <math>n=4</math> or <math>n</math> prime(by Wilson's Theorem). Thus, there are <math>16</math> integers that do not work, which means the answer is <math>\boxed{\mathbf{(D)}\ 34}</math>. | + | is not an integer if <math>n^2 \nmid n!</math>. This happens only for <math>n=4</math> or <math>n</math> prime (by Wilson's Theorem). Thus, there are <math>16</math> integers that do not work, which means the answer is <math>\boxed{\mathbf{(D)}\ 34}</math>. |
==See Also== | ==See Also== | ||
Revision as of 20:46, 17 February 2019
- The following problem is from both the 2019 AMC 10A #25 and 2019 AMC 12A #24, so both problems redirect to this page.
Problem
For how many integers 
between 
and 
, inclusive, is ![\[\frac{(n^2-1)!}{(n!)^n}\]](http://latex.artofproblemsolving.com/2/f/b/2fb0a7fcc7e8ea99426bae70e641eac53ea8d7e6.png)
an integer? (Recall that 
.)
Solution
The main insight is that
![\[\frac{(n^2)!}{(n!)^{n+1}}\]](http://latex.artofproblemsolving.com/e/4/5/e45e8bb08e31a5d89f1542589cbc346b1949bf8d.png)
is always an integer. This is true because it is precisely the number of ways to split up 
objects into unordered groups of size , which has to be an integer. Thus,
![\[\frac{(n^2)!}{(n!)^{n+1}}=\frac{n^2}{n!}\cdot\frac{(n^2-1)!}{(n!)^n}\]](http://latex.artofproblemsolving.com/6/5/f/65f8f0e8658d2574a4e834e69c8e586c74ec479a.png)
is not an integer if 
. This happens only for 
or prime (by Wilson's Theorem). Thus, there are 
integers that do not work, which means the answer is 
.
See Also
| 2019 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Last Problem | |
| 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 | ||
| 2019 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 23 |
Followed by Problem 25 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
