2020 AMC 8 Problems/Problem 12

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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

Notice that $5!$ = $2*3*4*5,$ and we can combine the numbers to create a larger factorial. To turn $9!$ into $10!,$ we need to multiply $9!$ by $2*5,$ which equals to $10!.$

Therefore, we have

\[10!*12=12*N!.\] We can cancel the $12$'s, since we are multiplying them on both sides of the equation.

We have

\[10!=N!.\] From here, it is obvious that $N=\boxed{10\textbf{(A)}}.$

-iiRishabii

Solution 2

$5!\cdot 9!=12\cdot N!$
$120\cdot 9!=12\cdot N!$
$12\cdot 10\cdot 9!=12\cdot N!$
$12 \cdot 10!=12\cdot N!$
$N=10 \implies\boxed{\textbf{(A) }10}$.
~junaidmansuri

Solution 3 (Non-rigorous)

We can see that the answers B through E have the factor 11, but there is no 11 in $5!\cdot9!$. Therefore, the answer must be the only answer without a $11$ factor, $A$.

~Windigo

Solution 4

Notice that $5!\cdot 9!=12\cdot 10\cdot 9!=12\cdot 10!$. We are also told that $12\cdot 10!=12*N!$ from where it is obvious that $N=\textbf{(A)}10$.

-franzliszt

Solution 5

We see that $5!\cdot9! = 5\cdot4\cdot3\cdot2\cdot1\cdot9! = 12\cdot{N!}$. Notice that $12 = 3\cdot4$, so: \[5\cdot2\cdot1\cdot9! = N!\] We see that $5\cdot2\cdot1\cdot9! = 10\cdot9! = 10! = N!$. So $N = \boxed{10} = \textbf{(A)}10$.

Solution 6

We note that $5!\cdot 9!=12\cdot 10\cdot 9!=12\cdot 10!$ we can actually get 120*9!= 12*N! which then you just get to your conclusion 10! which is equal to answer choice $N=\textbf{(A)}10$.

Video Solution

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

~savannahsolver

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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