2020 AMC 8 Problems/Problem 12

Revision as of 06:10, 18 November 2020 by Jmansuri (talk | contribs) (Solution 2)

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

See also

2020 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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