Difference between revisions of "2019 AMC 10A Problems/Problem 25"
m (changed headers) |
(→Solution 3 (Fakesolve)) |
||
(34 intermediate revisions by 16 users not shown) | |||
Line 7: | Line 7: | ||
<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 1== | |
− | |||
The main insight is that | The main insight is that | ||
Line 17: | Line 16: | ||
<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> | + | is an integer if <math>n^2 \mid n!</math>, or in other words, if <math>\frac{(n-1)!}{n}</math>, is an integer. 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 <math>15 + 1 = 16</math> terms for which |
<cmath>\frac{(n^2-1)!}{(n!)^{n}}</cmath> | <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{\ | + | 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 <math>2</math> 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 <math>16</math> values of n make the expression not an integer and the answer is <math>50-16=\boxed{\textbf{(D)}\ 34}</math>. |
− | ===Solution 2 | + | SideNote: Another method to prove that <cmath>\frac{(n^2)!}{(n!)^{n+1}}</cmath> is always an integer is instead as follows using Number Theory. Notice that <math>n</math> will divide the numerator <math>n+1</math> times, since <math>n^2 = n \times n</math> contains not one but two factors of <math>n.</math> Also, for <math>a < n,</math> notice that <math>a</math> divides <math>(n^2)!</math> at least <cmath>\lfloor \frac{n^2}{a} \rfloor \ge \lfloor \frac{n^2}{n-1} \rfloor \ge \lfloor \frac{n^2 - 1}{n-1} \rfloor \ge n+1</cmath> times. Thus, each integer from <math>1</math> to <math>n</math> will divide <math>(n^2)!</math> at least <math>n+1</math> times, which proves such a lemma. <math>\square{}</math> |
− | 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 : | + | |
+ | ==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 <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> | <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 | + | 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> . | We also know that , <math>v_p (m^n) = n \cdot v_p (m)</math> . | ||
Line 47: | Line 48: | ||
Now we try the same for <math>n=p^2</math> , where p is a prime. By Legendre we arrive at: | Now we try the same for <math>n=p^2</math> , 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> | + | <cmath>v_p ((p^4 -1)!) = p^3 + p^2 + p -3</cmath> |
+ | |||
+ | (as <math>v_p(p^4!) = p^3 + p^2 + p + 1</math> and <math>p^4</math> contains 4 factors of <math>p</math>) and | ||
+ | |||
+ | <cmath>p^2 \cdot v_p (p^2 !) = p^3 + p^2 </cmath> | ||
Then we get: | 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 <math>i</math> is an integer greater than 2, satisfies the inequality :<cmath> n \cdot v_p (n!) \le v_p ((n^2 -1 )!)</cmath> | + | <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 <math>i</math> 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. | Therefore, there are 16 values that don't work and <math> 50-16 = \boxed{\mathbf{(D)}\ 34}</math> values that work. | ||
~qwertysri987 | ~qwertysri987 | ||
+ | |||
+ | ==Solution 3 (Non-Rigorous)== | ||
+ | Notice all <math>15</math> primes don't work as there are <math>n</math> factors of <math>n</math> in the denominator and <math>n-1</math> factors of <math>n</math> in the numerator. Further experimentation finds that <math>n=4</math> 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 <math>n</math> work. So we get <math>50-15-1=\boxed{34}</math> and that happens to be right. | ||
+ | |||
+ | ~[[BS2012]] | ||
==See Also== | ==See Also== | ||
+ | Video Solution by Richard Rusczyk: | ||
+ | https://www.youtube.com/watch?v=9klaWnZojq0 | ||
{{AMC10 box|year=2019|ab=A|num-b=24|after=Last Problem}} | {{AMC10 box|year=2019|ab=A|num-b=24|after=Last Problem}} | ||
{{AMC12 box|year=2019|ab=A|num-b=23|num-a=25}} | {{AMC12 box|year=2019|ab=A|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 15:21, 11 November 2023
- 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) | ||
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.