Difference between revisions of "2020 AMC 8 Problems/Problem 12"
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==Problem== | ==Problem== | ||
− | For a positive integer <math>n</math>, the factorial notation <math>n!</math> represents the product of the integers from <math>n</math> to <math>1</math>. | + | For a positive integer <math>n</math>, the factorial notation <math>n!</math> represents the product of the integers from <math>n</math> to <math>1</math>. What value of <math>N</math> satisfies the following equation? <cmath>5!\cdot 9!=12\cdot N!</cmath> |
<math>\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14\qquad</math> | <math>\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14\qquad</math> | ||
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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>. | ||
− | ==Solution 3 (using answer choices)== | + | ==Solution 3 (using answer choices and elimination)== |
− | 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 answer must be <math>\boxed{\textbf{(A) }10}</math>. | + | 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>. The factor 11 is in every answer choice after <math>\boxed{\textbf{(A) }10}</math>, so four of the possible answers are eliminated. Therefore, the answer must be <math>\boxed{\textbf{(A) }10}</math>. |
+ | ~edited by HW73 | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | We notice that <math>5! \cdot 9! = (5!)^2 \cdot (9 \cdot 8 \cdot 7 \cdot 6).</math> | ||
+ | |||
+ | We know that <math>5! = 120,</math> so we have <math>120(5! \cdot 9 \cdot 8 \cdot 7 \cdot 6) = 12 \cdot N!</math> | ||
+ | |||
+ | Isolating <math>N!</math> we have <math>N! = 10 \cdot 5! \cdot 9 \cdot 8 \cdot 7 \cdot 6 \Rightarrow N! = 10! \Rightarrow N = \boxed{\textbf{(A) }10}.</math> | ||
+ | |||
+ | ~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== | ||
+ | https://youtu.be/9k59v-Fr3aE | ||
+ | |||
+ | ~savannahsolver | ||
==Video Solution== | ==Video Solution== | ||
− | https://youtu.be/ | + | https://youtu.be/xjwDsaRE_Wo |
+ | |||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/YnwkBZTv5Fw?t=504 | ||
+ | |||
+ | ~Interstigation | ||
==See also== | ==See also== | ||
{{AMC8 box|year=2020|num-b=11|num-a=13}} | {{AMC8 box|year=2020|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 00:07, 24 April 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.