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

Revision as of 07:50, 20 November 2020

Problem

For a positive integer $n$ , the factorial notation $n!$ represents the product of the integers from $n$ to $1$ . (For example, $6! = 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$ .) What value of $N$ satisfies the following equation? $5!\cdot 9!=12\cdot N!$

$\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14\qquad$

Solution 1

We have $5! = 2 \cdot 3 \cdot 4 \cdot 5$ , and $2 \cdot 5 \cdot 9! = 10 \cdot 9! = 10!$ . Therefore the equation becomes $3 \cdot 4 \cdot 10! = 12 \cdot N!$ , and so $12 \cdot 10! = 12 \cdot N!$ . Cancelling the $12$ s, it is clear that $N=\boxed{\textbf{(A) }10}$ .

Solution 2 (variation of Solution 1)

Since $5! = 120$ , we obtain $120\cdot 9!=12\cdot N!$ , which becomes $12\cdot 10\cdot 9!=12\cdot N!$ and thus $12 \cdot 10!=12\cdot N!$ . We therefore deduce $N=\boxed{\textbf{(A) }10}$ .

Solution 3 (using answer choices)

We can see that the answers $\textbf{(B)}$ to $\textbf{(E)}$ contain a factor of $11$ , but there is no such factor of $11$ in $5! \cdot 9!$ . Therefore, the answer must be $\boxed{\textbf{(A) }10}$ .

Video Solution

https://youtu.be/9k59v-Fr3aE

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 1: / Line 1: @@
-For a positive integer <math>n</math>, the factorial notation <math>n!</math> represents 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>
+==Problem==
+For a positive integer <math>n</math>, the factorial notation <math>n!</math> represents the product of the integers from <math>n</math> to <math>1</math>. (For example, <math>6! = 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1</math>.) What value of <math>N</math> satisfies the following equation? <cmath>5!\cdot 9!=12\cdot N!</cmath>
 <math>\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14\qquad</math>
 ==Solution 1==
+We have <math>5! = 2 \cdot 3 \cdot 4 \cdot 5</math>, and <math>2 \cdot 5 \cdot 9! = 10 \cdot 9! = 10!</math>. Therefore the equation becomes <math>3 \cdot 4 \cdot 10! = 12 \cdot N!</math>, and so <math>12 \cdot 10! = 12 \cdot N!</math>. Cancelling the <math>12</math>s, it is clear that <math>N=\boxed{\textbf{(A) }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>
+==Solution 2 (variation of Solution 1)==
+Since <math>5! = 120</math>, we obtain <math>120\cdot 9!=12\cdot N!</math>, which becomes <math>12\cdot 10\cdot 9!=12\cdot N!</math> and thus <math>12 \cdot 10!=12\cdot N!</math>. We therefore deduce <math>N=\boxed{\textbf{(A) }10}</math>.
-Therefore, we have
+==Solution 3 (using answer choices)==
+We can see that the answers <math>\textbf{(B)}</math> to <math>\textbf{(E)}</math> contain a factor of <math>11</math>, but there is no such factor of <math>11</math> in <math>5! \cdot 9!</math>. Therefore, the answer must be <math>\boxed{\textbf{(A) }10}</math>.
-<cmath>10!*12=12*N!.</cmath>
-We can cancel the <math>12</math>'s, since we are multiplying them on both sides of the equation.
-We have
-<cmath>10!=N!.</cmath>
-From here, it is obvious that <math>N=\boxed{10\textbf{(A)}}.</math>
--iiRishabii
-==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>
-~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri]
-==Solution 3 (Non-rigorous)==
-We can see that the answers B through E have the factor 11, but there is no 11 in <math>5!\cdot9!</math>. Therefore, the answer must be the only answer without a <math>11</math> factor, <math>A</math>.
-~Windigo
-==Solution 4==
-Notice that <math>5!\cdot 9!=12\cdot 10\cdot 9!=12\cdot 10!</math>. We are also told that <math>12\cdot 10!=12*N!</math> from where it is obvious that <math>N=\textbf{(A)}10</math>.
--franzliszt
-==Solution 5==
-We see that <math>5!\cdot9! = 5\cdot4\cdot3\cdot2\cdot1\cdot9! = 12\cdot{N!}</math>. Notice that <math>12 = 3\cdot4</math>, so:
-<cmath>5\cdot2\cdot1\cdot9! = N!</cmath>
-We see that <math>5\cdot2\cdot1\cdot9! = 10\cdot9! = 10! = N!</math>. So <math>N = \boxed{10} = \textbf{(A)}10</math>.
-==Solution 6==
-We note that <math>5!\cdot 9!=12\cdot 10\cdot 9!=12\cdot 10!</math> we can actually get 120*9!= 12*N! which then you just get to your conclusion 10! which is equal to answer choice <math>N=\textbf{(A)}10</math>.
 ==Video Solution==
 https://youtu.be/9k59v-Fr3aE
-~savannahsolver
 ==See also==
 {{AMC8 box|year=2020|num-b=11|num-a=13}}
 {{MAA Notice}}

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