Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2003 AMC 12A Problems/Problem 23"

(→‎Problem 23)
m
 
(27 intermediate revisions by 12 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
 +
How many perfect squares are divisors of the product <math>1! \cdot 2! \cdot 3! \cdot \hdots \cdot 9!</math>?
  
−
If <math>a\geq b > 1,</math> what is the largest possible value of <math>\log_{a}(a/b) + \log_{b}(b/a)?</math>
+
<math> \textbf{(A)}\ 504\qquad\textbf{(B)}\ 672\qquad\textbf{(C)}\ 864\qquad\textbf{(D)}\ 936\qquad\textbf{(E)}\ 1008 </math>
  
−
<math>
+
== Solution 1 ==
−
\mathrm{(A)}\ -2      \qquad
 
−
\mathrm{(B)}\ 0    \qquad
 
−
\mathrm{(C)}\ 2      \qquad
 
−
\mathrm{(D)}\ 3      \qquad
 
−
\mathrm{(E)}\ 4
 
−
</math>
 
  
−
== Solution ==
+
We want to find the number of perfect square factors in the product of all the factorials of numbers from <math>1 - 9</math>. We can write this out and take out the factorials, and then find a prime factorization of the entire product. We can also find this prime factorization by finding the number of times each factor is repeated in each factorial. This comes out to be equal to <math>2^{30} \cdot 3^{13} \cdot 5^5 \cdot 7^3</math>. To find the amount of perfect square factors, we realize that each exponent in the prime factorization must be even: <math>2^{15} \cdot 3^{6}\cdot 5^2 \cdot 7^1</math>. To find the total number of possibilities, we add <math>1</math> to each exponent and multiply them all together. This gives us <math>16 \cdot 7 \cdot 3 \cdot 2 = 672</math> <math>\Rightarrow\boxed{\mathrm{(B)}}</math>.
  
−
Using logarithmic rules, we see that
+
== Solution 2 ==
 +
(Explanation of 1)
  
−
<cmath>\log_{a}a-\log_{a}b+\log_{b}b-\log_{b}a = 2-(\log_{a}b+\log_{b}a</cmath>
+
Factorials up to 9 in product lead to prime factorization <math>2^{30}</math> <math>*</math> <math>3^{13}</math> <math>*</math> <math>5^5</math> <math>*</math> <math>7^3</math> So the number of pairs possible is { <math>2^{0}</math> <math>,</math> <math>2^{2}</math> <math>,</math> ... <math>2^{30}</math> }{ <math>3^{0}</math> <math>,</math> <math>3^{2}</math> <math>,</math> ... <math>3^{12}</math> }{ <math>5^{0}</math> <math>,</math> <math>5^{2}</math> <math>,</math> ... <math>5^{4}</math> }{ <math>7^{0}</math> <math>,</math> <math>7^{2}</math> }
−
<cmath>=2-(\log_{a}b+\frac {1}{\log_{a}b})</cmath>
 
  
−
Since <math>a</math> and <math>b</math> are both positive, using [[AM-GM]] gives that the term in parentheses must be at least <math>2</math>, so the largest possible values is <math>2-2=\boxed{0}.</math>
+
Which leads to resulting number of pairs = <math>16*7*3*2</math> <math>=</math> <math>672</math> <math>\Rightarrow\boxed{\mathrm{(B)}}</math>.
 +
 
 +
== See Also ==
 +
{{AMC12 box|year=2003|ab=A|num-b=22|num-a=24}}
 +
 
 +
[[Category:Intermediate Algebra Problems [Introductory Number Theory Problems]]
 +
{{MAA Notice}}

Latest revision as of 20:03, 29 May 2023

Problem

How many perfect squares are divisors of the product $1! \cdot 2! \cdot 3! \cdot \hdots \cdot 9!$?

$\textbf{(A)}\ 504\qquad\textbf{(B)}\ 672\qquad\textbf{(C)}\ 864\qquad\textbf{(D)}\ 936\qquad\textbf{(E)}\ 1008$

Solution 1

We want to find the number of perfect square factors in the product of all the factorials of numbers from $1 - 9$. We can write this out and take out the factorials, and then find a prime factorization of the entire product. We can also find this prime factorization by finding the number of times each factor is repeated in each factorial. This comes out to be equal to $2^{30} \cdot 3^{13} \cdot 5^5 \cdot 7^3$. To find the amount of perfect square factors, we realize that each exponent in the prime factorization must be even: $2^{15} \cdot 3^{6}\cdot 5^2 \cdot 7^1$. To find the total number of possibilities, we add $1$ to each exponent and multiply them all together. This gives us $16 \cdot 7 \cdot 3 \cdot 2 = 672$ $\Rightarrow\boxed{\mathrm{(B)}}$.

Solution 2

(Explanation of 1)

Factorials up to 9 in product lead to prime factorization $2^{30}$ $*$ $3^{13}$ $*$ $5^5$ $*$ $7^3$ So the number of pairs possible is { $2^{0}$ $,$ $2^{2}$ $,$ ... $2^{30}$ }{ $3^{0}$ $,$ $3^{2}$ $,$ ... $3^{12}$ }{ $5^{0}$ $,$ $5^{2}$ $,$ ... $5^{4}$ }{ $7^{0}$ $,$ $7^{2}$ }

Which leads to resulting number of pairs = $16*7*3*2$ $=$ $672$ $\Rightarrow\boxed{\mathrm{(B)}}$.

See Also

2003 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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

[[Category:Intermediate Algebra Problems [Introductory Number Theory Problems]] The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_23&oldid=193698"