Difference between revisions of "2020 AMC 8 Problems/Problem 12"
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We have <math>5! = 2 \cdot 3 \cdot 4 \cdot 5</math>, and <math>2 \cdot 5 \cdot 9! = 10 \cdot 9! = 10!</math>. Therefore the equation becomes <math>3 \cdot 4 \cdot 10! = 12 \cdot N!</math>, and so <math>12 \cdot 10! = 12 \cdot N!</math>. Cancelling the <math>12</math>s, it is clear that <math>N=\boxed{\textbf{(A) }10}</math>. | We have <math>5! = 2 \cdot 3 \cdot 4 \cdot 5</math>, and <math>2 \cdot 5 \cdot 9! = 10 \cdot 9! = 10!</math>. Therefore the equation becomes <math>3 \cdot 4 \cdot 10! = 12 \cdot N!</math>, and so <math>12 \cdot 10! = 12 \cdot N!</math>. Cancelling the <math>12</math>s, it is clear that <math>N=\boxed{\textbf{(A) }10}</math>. | ||
| − | ==Solution 2 ( | + | ==Solution 2 (variant of Solution 1)== |
Since <math>5! = 120</math>, we obtain <math>120\cdot 9!=12\cdot N!</math>, which becomes <math>12\cdot 10\cdot 9!=12\cdot N!</math> and thus <math>12 \cdot 10!=12\cdot N!</math>. We therefore deduce <math>N=\boxed{\textbf{(A) }10}</math>. | Since <math>5! = 120</math>, we obtain <math>120\cdot 9!=12\cdot N!</math>, which becomes <math>12\cdot 10\cdot 9!=12\cdot N!</math> and thus <math>12 \cdot 10!=12\cdot N!</math>. We therefore deduce <math>N=\boxed{\textbf{(A) }10}</math>. | ||
Revision as of 07:55, 20 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? ![\[5!\cdot 9!=12\cdot N!\]](http://latex.artofproblemsolving.com/7/a/a/7aaf22ceff79c5a5063175866cf2cba2e4a57f9b.png)
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 answer 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. 
