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

(Removed duplicate solutions and improved LaTeX, grammar, and clarity)
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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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==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>. (For example, <math>6! = 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 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>
  
 
==Solution 1==
 
==Solution 1==
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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>.
  
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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==Solution 2 (variation of Solution 1)==
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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>.
  
Therefore, we have
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==Solution 3 (using answer choices)==
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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>. Therefore, the answer must be <math>\boxed{\textbf{(A) }10}</math>.
  
<cmath>10!*12=12*N!.</cmath>
 
We can cancel the <math>12</math>'s, since we are multiplying them on both sides of the equation.
 
 
We have
 
 
<cmath>10!=N!.</cmath>
 
From here, it is obvious that <math>N=\boxed{10\textbf{(A)}}.</math>
 
 
-iiRishabii
 
 
==Solution 2==
 
<math>5!\cdot 9!=12\cdot N!</math><br><math>120\cdot 9!=12\cdot N!</math><br><math>12\cdot 10\cdot 9!=12\cdot N!</math><br><math>12 \cdot 10!=12\cdot N!</math><br><math>N=10 \implies\boxed{\textbf{(A) }10}</math>.<br>
 
~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri]
 
 
==Solution 3 (Non-rigorous)==
 
We can see that the answers B through E have the factor 11, but there is no 11 in <math>5!\cdot9!</math>. Therefore, the answer must be the only answer without a <math>11</math> factor, <math>A</math>.
 
 
~Windigo
 
 
==Solution 4==
 
 
Notice that <math>5!\cdot 9!=12\cdot 10\cdot 9!=12\cdot 10!</math>. We are also told that <math>12\cdot 10!=12*N!</math> from where it is obvious that <math>N=\textbf{(A)}10</math>.
 
 
-franzliszt
 
 
==Solution 5==
 
We see that <math>5!\cdot9! = 5\cdot4\cdot3\cdot2\cdot1\cdot9! = 12\cdot{N!}</math>. Notice that <math>12 = 3\cdot4</math>, so:
 
<cmath>5\cdot2\cdot1\cdot9! = N!</cmath>
 
We see that <math>5\cdot2\cdot1\cdot9! = 10\cdot9! = 10! = N!</math>. So <math>N = \boxed{10} = \textbf{(A)}10</math>.
 
==Solution 6==
 
We note that <math>5!\cdot 9!=12\cdot 10\cdot 9!=12\cdot 10!</math> we can actually get 120*9!= 12*N! which then you just get to your conclusion 10! which is equal to answer choice <math>N=\textbf{(A)}10</math>.
 
 
==Video Solution==
 
==Video Solution==
 
https://youtu.be/9k59v-Fr3aE
 
https://youtu.be/9k59v-Fr3aE
 
~savannahsolver
 
  
 
==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}}

Revision as of 08:50, 20 November 2020

Problem

For a positive integer $n$, the factorial notation $n!$ represents the product of the integers from $n$ to $1$. (For example, $6! = 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$.) 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 (variation 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)

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 $5! \cdot 9!$. Therefore, the answer must be $\boxed{\textbf{(A) }10}$.

Video Solution

https://youtu.be/9k59v-Fr3aE

See also

2020 AMC 8 (ProblemsAnswer KeyResources)
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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