Difference between revisions of "2003 AMC 12A Problems/Problem 23"
Cricketswift (talk | contribs) m (→Problem) |
Captainsnake (talk | contribs) |
||
Line 1: | Line 1: | ||
− | of the product <math>1! \cdot 2! \cdot 3! \cdot \hdots \cdot 9!</math>? | + | 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> | <math> \textbf{(A)}\ 504\qquad\textbf{(B)}\ 672\qquad\textbf{(C)}\ 864\qquad\textbf{(D)}\ 936\qquad\textbf{(E)}\ 1008 </math> |
Revision as of 11:53, 5 November 2020
How many perfect squares are divisors of the product ?
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.