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? $5!\cdot 9!=12\cdot N!$

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

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

2020 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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2020 AMC 8 Problems/Problem 12

Solution 1

Solution 2

See also