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

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Solution 1
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For positive integers <math>n</math>, the notation <math>n!</math> denotes the product of the integers from <math>n</math> to <math>1</math>. What value of <math>N</math> satisfies the following equation? <cmath>5!\cdot 9!=12\cdot N!</cmath>
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==Solution 1==
  
 
Notice that <math>5!</math> = <math>2*3*4*5,</math> and we can combine the numbers to create a larger factorial. To turn <math>9!</math> into <math>10!,</math> we need to multiply <math>9!</math> by <math>2*5,</math> which equals to <math>10!.</math>  
 
Notice that <math>5!</math> = <math>2*3*4*5,</math> and we can combine the numbers to create a larger factorial. To turn <math>9!</math> into <math>10!,</math> we need to multiply <math>9!</math> by <math>2*5,</math> which equals to <math>10!.</math>  

Revision as of 00:58, 18 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 $12s,$ since we are multiplying them on both sides of the equation.

We have

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

-iiRishabii