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

Revision as of 00:51, 30 November 2021

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)

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 $\boxed{\textbf{(A) }10}$ .

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 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

@@ Line 12: / Line 12: @@
 ==Solution 3 (using answer choices)==
 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>.
+==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==

2020 AMC 8 (Problems • Answer Key • Resources)
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