Difference between revisions of "2019 AMC 10A Problems/Problem 25"
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<math>\textbf{(A) } 31 \qquad \textbf{(B) } 32 \qquad \textbf{(C) } 33 \qquad \textbf{(D) } 34 \qquad \textbf{(E) } 35</math> | <math>\textbf{(A) } 31 \qquad \textbf{(B) } 32 \qquad \textbf{(C) } 33 \qquad \textbf{(D) } 34 \qquad \textbf{(E) } 35</math> | ||
− | ==Solution== | + | ==Solution 1== |
The main insight is that | The main insight is that | ||
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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>. | 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>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | We can use the P-Adic Valuation of n to solve this problem (recall the P-Adic Valuation of 'n' is denoted by <math>v_p (n)</math> and is defined as the greatest power of some prime 'p' that divides n. For example, <math>v_2 (6)=1</math> or <math>v_7 (245)=2</math> .) Using Legendre's formula, we know that : | ||
+ | |||
+ | <cmath> v_p (n!)= \sum_{i=1}^\infty \lfloor \frac {n}{p^i} \rfloor </cmath> | ||
+ | |||
+ | 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 , <math>v_p (m^n) = n \cdot v_p (m)</math> . | ||
+ | Knowing that <math>a\mid b</math> iff <math>v_p (a) \le v_p (b)</math> , we have that : | ||
+ | |||
+ | <cmath> n \cdot v_p (n!) \le v_p ((n^2 -1 )!) </cmath> and we must find all n for which this is true. | ||
+ | |||
+ | If we plug in <math>n=p</math>, by Legendre's we get two equations: | ||
+ | |||
+ | <cmath> v_p ((n^2 -1)!) = \sum_{i=1}^\infty \lfloor \frac {n^2 -1}{p^i} \rfloor = (p-1)+0+...+0 = p-1 </cmath> | ||
+ | |||
+ | And we also get : | ||
+ | |||
+ | <cmath> v_p ((n!)^n) = n \cdot v_p (n!)= n \cdot \sum_{i=1}^\infty \lfloor \frac {n}{p^i} \rfloor = p \cdot ( 1+0+...0) = p </cmath> | ||
+ | |||
+ | But we are asked to prove that <math> n \cdot v_p (n!) \le v_p ((n^2 -1 )!) \Longrightarrow p \le p-1 </math> which is false for all 'n' where n is prime. | ||
+ | |||
+ | Now we try the same for n=p^2 , where p is a prime. By Legendre we arrive at: | ||
+ | |||
+ | <cmath>v_p ((p^4 -1)!) = p^3 + p^2 + p -3</cmath> and <cmath>p^2 \cdot v_p (p^2 !) = p^3 + p^2 </cmath> | ||
+ | |||
+ | Then we get: | ||
+ | |||
+ | <cmath> p^2 \cdot v_p (p!) \le v_p ((n^4 -1)!) \Longrightarrow p^3 + p^2 \le p^3 + p^2 + p -3 </cmath> Which is true for all primes except for 2, so <math>2^2 = 4</math> doesn't work. It can easily be verified that for all <math>n=p^i</math> where i is an integer greater than 2, satisfies the inequality :<cmath> n \cdot v_p (n!) \le v_p ((n^2 -1 )!)</cmath>. | ||
+ | |||
+ | Therefore, there are 16 values that don't work and <math> 50-16 = \boxed{\mathbf{(D)}\ 34}</math> values that work. | ||
+ | |||
+ | ~qwertysri987 | ||
==See Also== | ==See Also== |
Revision as of 12:51, 28 June 2019
- The following problem is from both the 2019 AMC 10A #25 and 2019 AMC 12A #24, so both problems redirect to this page.
Contents
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 . 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 .
Solution 2
We can use the P-Adic Valuation 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 iff , 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 n=p^2 , where p is a prime. By Legendre we arrive at:
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 i is an integer greater than 2, satisfies the inequality :.
Therefore, there are 16 values that don't work and values that work.
~qwertysri987
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.