Difference between revisions of "2003 AMC 12A Problems/Problem 23"
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+ | == Problem == | ||
+ | How many perfect squares are divisors of the product <math>1! \cdot 2! \cdot 3! \cdot \hdots \cdot 9!</math>? | ||
− | + | <math> \textbf{(A)}\ 504\qquad\textbf{(B)}\ 672\qquad\textbf{(C)}\ 864\qquad\textbf{(D)}\ 936\qquad\textbf{(E)}\ 1008 </math> | |
− | ==Solution== | + | == Solutions == |
+ | === Solution 1 === | ||
+ | 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>. | ||
− | We | + | === Solution 2 === |
+ | ( Just Explanation Of 1 ) | ||
+ | |||
+ | 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> } | ||
+ | |||
+ | 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 == | == See Also == | ||
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 12:59, 19 January 2021
Problem
How many perfect squares are divisors of the product ?
Solutions
Solution 1
We want to find the number of perfect square factors in the product of all the factorials of numbers from . 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 . To find the amount of perfect square factors, we realize that each exponent in the prime factorization must be even: . To find the total number of possibilities, we add to each exponent and multiply them all together. This gives us .
Solution 2
( Just Explanation Of 1 )
We can easily find that factorials upto 9 in product lead to prime factorization So number of pairs possible = { ... }{ ... }{ ... }{ }
Which leads to resulting number of pairs = .
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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