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

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==Solution 2==
 
==Solution 2==
 
<math>5!\cdot 9!=12\cdot N!</math><br><math>120\cdot 9!=12\cdot N!</math><br><math>12\cdot 10\cdot 9!=12\cdot N!</math><br><math>12 \cdot 10!=12\cdot N!</math><br><math>N=10 \implies\boxed{\textbf{(A) }10}</math>.<br>
 
<math>5!\cdot 9!=12\cdot N!</math><br><math>120\cdot 9!=12\cdot N!</math><br><math>12\cdot 10\cdot 9!=12\cdot N!</math><br><math>12 \cdot 10!=12\cdot N!</math><br><math>N=10 \implies\boxed{\textbf{(A) }10}</math>.<br>
−
~jmansuri
+
~ junaidmansuri
  
 
==See also==
 
==See also==
 
{{AMC8 box|year=2020|num-b=11|num-a=13}}
 
{{AMC8 box|year=2020|num-b=11|num-a=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 06:26, 18 November 2020

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

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

See also

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

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