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Difference between revisions of "2003 AMC 12A Problems/Problem 23"
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== Problem ==
 
== Problem ==
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How many perfect squares are divisors of the product <math>1! \cdot 2! \cdot 3! \cdot \hdots \cdot 9!</math>?
  
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If <math>a\geq b > 1,</math> what is the largest possible value of <math>\log_{a}(a/b) + \log_{b}(b/a)?</math>
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<math> \textbf{(A)}\ 504\qquad\textbf{(B)}\ 672\qquad\textbf{(C)}\ 864\qquad\textbf{(D)}\ 936\qquad\textbf{(E)}\ 1008 </math>
  
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<math>
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== Solutions ==
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\mathrm{(A)}\ -2     \qquad
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=== Solution 1 ===
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\mathrm{(B)}\ 0    \qquad
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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>.
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\mathrm{(C)}\ 2     \qquad
 
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\mathrm{(D)}\ 3     \qquad
 
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\mathrm{(E)}\ 4
 
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</math>
 
  
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== Solution ==
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=== Solution 2 ===
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( Just Explanation Of 1 )
  
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Using logarithmic rules, we see that
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We can easily find that factorials upto 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 number of pairs possible = { <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> }
  
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<cmath>\log_{a}a-\log_{a}b+\log_{b}b-\log_{b}a = 2-(\log_{a}b+\log_{b}a</cmath>
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Which leads to resulting number of pairs = <math>16*7*3*2</math> <math>=</math> <math>672</math> <math>\Rightarrow\boxed{\mathrm{(B)}}</math>.
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<cmath>=2-(\log_{a}b+\frac {1}{\log_{a}b})</cmath>
 
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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>
 
  
 
== See Also ==
 
== See Also ==
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[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Revision as of 13:59, 19 January 2021

Problem

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

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

Solutions

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

( Just Explanation Of 1 )

We can easily find that factorials upto 9 in product lead to prime factorization $2^{30}$ $*$ $3^{13}$ $*$ $5^5$ $*$ $7^3$ So number of pairs possible = { $2^{0}$ $,$ $2^{2}$ $,$ ... $2^{30}$ }{ $3^{0}$ $,$ $3^{2}$ $,$ ... $3^{12}$ }{ $5^{0}$ $,$ $5^{2}$ $,$ ... $5^{4}$ }{ $7^{0}$ $,$ $7^{2}$ }

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

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