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

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==Solution==
 
==Solution==
  
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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. Anyway, this comes out to be equal to 2^30 * 3^13 * 5^5 * 7^3. To find the amount of perfect square factors, we realize that each exponent in the prime factorization must be even. This gives us 16*7*3*2 = 672 (B).
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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}</math> <math>*</math> <math>3^{13}</math> <math>*</math> <math>5^5</math> <math>*</math> <math>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}</math> <math>*</math> <math>3^{6}</math> <math>*</math> <math>5^2</math> <math>*</math> <math>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*7*3*2</math> <math>=</math> <math>672</math> <math>\Rightarrow\mathrm{(B)}</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 19:55, 25 December 2012

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

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 $2^{30}$ $*$ $3^{13}$ $*$ $5^5$ $*$ $7^3$. To find the amount of perfect square factors, we realize that each exponent in the prime factorization must be even: $2^{15}$ $*$ $3^{6}$ $*$ $5^2$ $*$ $7^1$. To find the total number of possibilities, we add $1$ to each exponent and multiply them all together. This gives us $16*7*3*2$ $=$ $672$ $\Rightarrow\mathrm{(B)}$.

See Also

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