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Difference between revisions of "2020 AMC 8 Problems/Problem 12"
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[b]Solution 1[\b]
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==Problem==
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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>
  
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Notice that <math>5!</math> = <math>2*3*4*5,</math> and we can combine the numbers to create a larger factorial. To turn <math>9!</math> into <math>10!,</math> we need to multiply <math>9!</math> by <math>2*5,</math> which equals to <math>10!.</math>  
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<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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Therefore, we have
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==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>.
  
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<cmath>10!*12=12*N!.</cmath>
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==Solution 2 (variant of Solution 1)==
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We can cancel the <math>12s,</math> since we are multiplying them on both sides of the equation.
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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>.
  
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We have
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==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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~edited by HW73
  
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<cmath>10!=N!.</cmath>
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==Solution 4==
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From here, it is obvious that <math>N=10(A).</math>
 
  
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-iiRishabii
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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
 +
 
 +
==Video Solution by Interstigation==
 +
https://youtu.be/YnwkBZTv5Fw?t=504
 +
 
 +
~Interstigation
 +
 
 +
==See also==
 +
{{AMC8 box|year=2020|num-b=11|num-a=13}}
 +
{{MAA Notice}}

Latest revision as of 15:34, 26 January 2024

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 after $\boxed{\textbf{(A) }10}$, 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

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

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