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)== | ||
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 NiuniuMaths (Easy to understand!)== | ||
+ | https://www.youtube.com/watch?v=bHNrBwwUCMI | ||
+ | |||
+ | ~NiuniuMaths | ||
+ | |||
+ | ==Video Solution by Math-X (First understand the problem!!!)== | ||
+ | https://youtu.be/UnVo6jZ3Wnk?si=aiVOk6HkZErYYi6P&t=1568 | ||
+ | |||
+ | ~Math-X | ||
+ | |||
+ | ==Video Solution (🚀 Fast 🚀)== | ||
+ | https://youtu.be/5LLNBB83bWA | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ==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== | ||
https://youtu.be/xjwDsaRE_Wo | 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 15:34, 26 January 2024
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2 (variant of Solution 1)
- 4 Solution 3 (using answer choices and elimination)
- 5 Solution 4
- 6 Video Solution by NiuniuMaths (Easy to understand!)
- 7 Video Solution by Math-X (First understand the problem!!!)
- 8 Video Solution (🚀 Fast 🚀)
- 9 Video Solution by North America Math Contest Go Go Go
- 10 Video Solution by WhyMath
- 11 Video Solution
- 12 Video Solution by Interstigation
- 13 See also
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 NiuniuMaths (Easy to understand!)
https://www.youtube.com/watch?v=bHNrBwwUCMI
~NiuniuMaths
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/UnVo6jZ3Wnk?si=aiVOk6HkZErYYi6P&t=1568
~Math-X
Video Solution (🚀 Fast 🚀)
~Education, the Study of Everything
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.