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>, or in other words, if <math>n \mid (n-1)!</math>. This condition is false precisely when <math>n=4</math> or <math>n</math> is prime, by Wilson's Theorem. There are <math>15</math> primes between <math>1</math> and <math>50</math>, inclusive, so the answer is <math>50- | + | is an integer if <math>n^2 \mid n!</math>, or in other words, if <math>n \mid (n-1)!</math>. This condition is false precisely when <math>n=4</math> or <math>n</math> is prime, by Wilson's Theorem. There are <math>15</math> primes between <math>1</math> and <math>50</math>, inclusive, so there are 15 + 1 = 16 terms for which |
+ | |||
+ | <cmath>\frac{(n^2-1)!}{(n!)^{n}}</cmath> | ||
+ | |||
+ | is potentially not an integer. It can be easily verified that the above expression is not an integer for <math>n=4</math> as there are more factors of 2 in the denominator than the numerator. Similarly, it can be verified that the above expression is not an integer for any prime <math>n=p</math>, as there are more factors of p in the denominator than the numerator. Thus all 16 values of n make the expression not an integer and the answer is <math>50-16=\boxed{\mathbf{(D)}\ 34}</math>. | ||
==See Also== | ==See Also== |
Revision as of 02:05, 2 March 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 , or in other words, if . This condition is false precisely when or is prime, by Wilson's Theorem. There are primes between and , inclusive, so there are 15 + 1 = 16 terms for which
is potentially not an integer. It can be easily verified that the above expression is not an integer for as there are more factors of 2 in the denominator than the numerator. Similarly, it can be verified that the above expression is not an integer for any prime , as there are more factors of p in the denominator than the numerator. Thus all 16 values of n make the expression not an integer and 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.