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Difference between revisions of "2020 AMC 8 Problems/Problem 12"

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==Solution 3 (using answer choices and elimination)==
 
==Solution 3 (using answer choices and elimination)==
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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>.
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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 except <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>.
 
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~edited by HW73
  
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~mathboy282
 
~mathboy282
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== Solution 5 ==
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Notice that <math>9!</math> is equivalent to <math>\frac{12!}{12 \cdot 11 \cdot 10}</math>. In the expression <math>\frac{12!}{12 \cdot 11 \cdot 10} \cdot 5 \cdot 4 \cdot 3 \cdot 2</math>, cancel out <math>12</math>, <math>4</math>, <math>3</math>, and <math>10</math>, <math>5</math>, <math>2</math>. The resulting equation is <math>\frac{12!}{11} = 12 \cdot N!</math>. The equation gives <math>12 \cdot 10! = 12 \cdot N!</math>. Therefore, N = 10, so the answer is <math>\boxed{(A)10}</math>.
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==Video Solution by NiuniuMaths (Easy to understand!)==
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https://www.youtube.com/watch?v=bHNrBwwUCMI
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~NiuniuMaths
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==Video Solution by Math-X (First understand the problem!!!)==
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https://youtu.be/UnVo6jZ3Wnk?si=aiVOk6HkZErYYi6P&t=1568
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~Math-X
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==Video Solution (🚀 Fast 🚀)==
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https://youtu.be/5LLNBB83bWA
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~Education, the Study of Everything
  
 
==Video Solution by North America Math Contest Go Go Go==
 
==Video Solution by North America Math Contest Go Go Go==
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{{AMC8 box|year=2020|num-b=11|num-a=13}}
 
{{AMC8 box|year=2020|num-b=11|num-a=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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[[Category:Introductory Algebra Problems]]

Latest revision as of 08:41, 6 June 2025

Problem

For a positive integer $n$, the factorial notation $n!$ represents the product of the integers from $n$ to $1$. What value of $N$ satisfies the following equation? \[5!\cdot 9!=12\cdot N!\]

$\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14\qquad$

Solution 1

We have $5! = 2 \cdot 3 \cdot 4 \cdot 5$, and $2 \cdot 5 \cdot 9! = 10 \cdot 9! = 10!$. Therefore, the equation becomes $3 \cdot 4 \cdot 10! = 12 \cdot N!$, and so $12 \cdot 10! = 12 \cdot N!$. Cancelling the $12$s, it is clear that $N=\boxed{\textbf{(A) }10}$.

Solution 2 (variant of Solution 1)

Since $5! = 120$, we obtain $120\cdot 9!=12\cdot N!$, which becomes $12\cdot 10\cdot 9!=12\cdot N!$ and thus $12 \cdot 10!=12\cdot N!$. We therefore deduce $N=\boxed{\textbf{(A) }10}$.

Solution 3 (using answer choices and elimination)

We can see that the answers $\textbf{(B)}$ to $\textbf{(E)}$ contain a factor of $11$, but there is no such factor of $11$ in $5! \cdot 9!$. The factor 11 is in every answer choice except $\boxed{\textbf{(A) }10}$, so four of the possible answers are eliminated. Therefore, the answer must be $\boxed{\textbf{(A) }10}$. ~edited by HW73

Solution 4

We notice that $5! \cdot 9! = (5!)^2 \cdot (9 \cdot 8 \cdot 7 \cdot 6).$

We know that $5! = 120,$ so we have $120(5! \cdot 9 \cdot 8 \cdot 7 \cdot 6) = 12 \cdot N!$

Isolating $N!$ we have $N! = 10 \cdot 5! \cdot 9 \cdot 8 \cdot 7 \cdot 6 \Rightarrow N! = 10! \Rightarrow N = \boxed{\textbf{(A) }10}.$

~mathboy282

Solution 5

Notice that $9!$ is equivalent to $\frac{12!}{12 \cdot 11 \cdot 10}$. In the expression $\frac{12!}{12 \cdot 11 \cdot 10} \cdot 5 \cdot 4 \cdot 3 \cdot 2$, cancel out $12$, $4$, $3$, and $10$, $5$, $2$. The resulting equation is $\frac{12!}{11} = 12 \cdot N!$. The equation gives $12 \cdot 10! = 12 \cdot N!$. Therefore, N = 10, so the answer is $\boxed{(A)10}$.

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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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