2019 AMC 10A Problems/Problem 25
- 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 1
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 , is an integer. This condition is false precisely when or is prime, by Wilson's Theorem. There are primes between and , inclusive, so there are 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 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 values of n make the expression not an integer and the answer is .
SideNote: Another method to prove that is always an integer is instead as follows using Number Theory. Notice that will divide the numerator times, since contains not one but two factors of Also, for notice that divides at least times. Thus, each integer from to will divide at least times, which proves such a lemma.
Solution 2
We can use the P-Adic Valuation (more info could be found here: Mathematicial notation) of n to solve this problem (recall the P-Adic Valuation of 'n' is denoted by and is defined as the greatest power of some prime 'p' that divides n. For example, or .) Using Legendre's formula, we know that :
Seeing factorials involved in the problem, this prompts us to use Legendre's Formula where n is a power of a prime.
We also know that , . Knowing that if , we have that :
and we must find all n for which this is true.
If we plug in , by Legendre's we get two equations:
And we also get :
But we are asked to prove that which is false for all 'n' where n is prime.
Now we try the same for , where p is a prime. By Legendre we arrive at:
(as and contains 4 factors of ) and
Then we get:
Which is true for all primes except for 2, so doesn't work. It can easily be verified that for all where is an integer greater than 2, satisfies the inequality :
Therefore, there are 16 values that don't work and values that work.
~qwertysri987
Solution 3 (Non-Rigorous)
Notice all primes don't work as there are factors of in the denominator and factors of in the numerator. Further experimentation finds that does not work as there are 11 factors of 2 in the numerator and 12 in the denominator. We also find that it seems that all other values of work. So we get and that happens to be right.
See Also
Video Solution by Richard Rusczyk: https://www.youtube.com/watch?v=9klaWnZojq0
2019 AMC 10A (Problems • Answer Key • Resources) | ||
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