Difference between revisions of "2020 AMC 8 Problems/Problem 12"
Sevenoptimus (talk | contribs) m (Minor fix) |
Mathboy2788 (talk | contribs) (→Solution 3 (using answer choices)) |
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==Solution 3 (using answer choices)== | ==Solution 3 (using answer choices)== | ||
− | We can see that the answers <math>\textbf{(B)}</math> to <math>\textbf{(E)}</math> contain a factor of <math>11</math>, but there is no such factor of <math>11</math> in <math>5! \cdot 9!</math>. Therefore, the | + | We can see that the answers <math>\textbf{(B)}</math> to <math>\textbf{(E)}</math> contain a factor of <math>11</math>, but there is no such factor of <math>11</math> in <math>5! \cdot 9!</math>. Therefore, the aanswer must be <math>\boxed{\textbf{(A) }10}</math>. |
==Video Solution== | ==Video Solution== |
Revision as of 22:37, 23 November 2020
Contents
Problem
For a positive integer , the factorial notation represents the product of the integers from to . (For example, .) What value of satisfies the following equation?
Solution 1
We have , and . Therefore the equation becomes , and so . Cancelling the s, it is clear that .
Solution 2 (variant of Solution 1)
Since , we obtain , which becomes and thus . We therefore deduce .
Solution 3 (using answer choices)
We can see that the answers to contain a factor of , but there is no such factor of in . Therefore, the aanswer must be .
Video Solution
See also
2020 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.