2020 AMC 8 Problems/Problem 12: Difference between revisions

Revision as of 07:41, 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

@@ Line 25: / Line 25: @@
 == Solution 5 ==
+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>.
 ==Video Solution by NiuniuMaths (Easy to understand!)==

2020 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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