Difference between revisions of "2019 AMC 10A Problems/Problem 25"
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<cmath>\frac{(n^2-1)!}{(n!)^n}=\frac{(n^2)!}{(n!)^{n+1}}\cdot\frac{n!}{n^2}</cmath> | <cmath>\frac{(n^2-1)!}{(n!)^n}=\frac{(n^2)!}{(n!)^{n+1}}\cdot\frac{n!}{n^2}</cmath> | ||
| − | is an integer if <math>n^2 \mid n!</math>, in other words if <math>n \mid (n-1)!</math>. This fails exactly for <math>n=4</math> or <math>n</math> prime (Wilson's Theorem). There are | + | is an integer if <math>n^2 \mid n!</math>, in other words if <math>n \mid (n-1)!</math>. This fails exactly for <math>n=4</math> or <math>n</math> prime (Wilson's Theorem). There are 15 primes between 1 and 50, inclusive, so the answer is <math>50-1-15=\boxed{\mathbf{(D)}\ 34}</math>. |
==See Also== | ==See Also== | ||
Revision as of 21:06, 24 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 . Thus,
![\[\frac{(n^2-1)!}{(n!)^n}=\frac{(n^2)!}{(n!)^{n+1}}\cdot\frac{n!}{n^2}\]](http://latex.artofproblemsolving.com/7/2/7/727e7d51f941f9e450a61f08646b5a1e5e270350.png)
is an integer if 
, in other words if 
. This fails exactly for 
or prime (Wilson's Theorem). There are 15 primes between 1 and 50, inclusive, so 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. 
