Difference between revisions of "2020 AMC 8 Problems/Problem 12"

Line 8: Line 8:
  
 
<cmath>10!*12=12*N!.</cmath>
 
<cmath>10!*12=12*N!.</cmath>
We can cancel the <math>12s,</math> since we are multiplying them on both sides of the equation.
+
We can cancel the <math>12</math>'s, since we are multiplying them on both sides of the equation.
  
 
We have
 
We have
  
 
<cmath>10!=N!.</cmath>
 
<cmath>10!=N!.</cmath>
From here, it is obvious that <math>N=10(A).</math>
+
From here, it is obvious that <math>N=\boxed{10\textbf{(A)}}.</math>
  
 
-iiRishabii
 
-iiRishabii

Revision as of 23:59, 17 November 2020

For positive integers $n$, the notation $n!$ denotes 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