2020 AMC 8 Problems/Problem 12

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

Video Solution

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

~savannahsolver

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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