Difference between revisions of "2019 AMC 10A Problems/Problem 25"
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<cmath>\frac{(n^2)!}{(n!)^{n+1}}</cmath> | <cmath>\frac{(n^2)!}{(n!)^{n+1}}</cmath> | ||
− | is always an integer. This is true because it is precisely the number of ways to split up <math>n^2</math> objects into <math>n</math> unordered groups of size <math>n</math> | + | is always an integer. This is true because it is precisely the number of ways to split up <math>n^2</math> objects into <math>n</math> unordered groups of size <math>n</math>. Thus, |
− | <cmath>\frac{(n^2)!}{(n!)^ | + | <cmath>\frac{(n^2-1)!}{(n!)^n}=\frac{(n^2)!}{(n!)^{n+1}}\cdot\frac{n!}{n^2}</cmath> |
− | is | + | 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 <math>15</math> 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 22:05, 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 an integer? (Recall that .)
Solution
The main insight is that
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,
is an integer if , in other words if . This fails exactly for or prime (Wilson's Theorem). There are 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.