Difference between revisions of "2020 AMC 8 Problems/Problem 12"
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==Solution 1== | ==Solution 1== | ||
− | 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 (variant of Solution 1)== | ==Solution 2 (variant of Solution 1)== |
Revision as of 10:47, 2 December 2022
Contents
Problem
For a positive integer , the factorial notation
represents the product of the integers from
to
. 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 and elimination)
We can see that the answers to
contain a factor of
, but there is no such factor of
in
. The factor 11 is in every answer choice after
, so four of the possible answers are eliminated. Therefore, the answer must be
.
~edited by HW73
Solution 4
We notice that
We know that so we have
Isolating we have
~mathboy282
Video Solution by North America Math Contest Go Go Go
https://www.youtube.com/watch?v=mYs1-Nbr0Ec
~North America Math Contest Go Go Go
Video Solution by WhyMath
~savannahsolver
Video Solution
Video Solution by Interstigation
https://youtu.be/YnwkBZTv5Fw?t=504
~Interstigation
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.