https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Mathboy2788&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T13:01:33ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_12&diff=1383262020 AMC 8 Problems/Problem 122020-11-24T02:37:17Z<p>Mathboy2788: /* Solution 3 (using answer choices) */</p>
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<div>==Problem==<br />
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><br />
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<math>\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14\qquad</math><br />
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==Solution 1==<br />
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>.<br />
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==Solution 2 (variant of Solution 1)==<br />
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>.<br />
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==Solution 3 (using answer choices)==<br />
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 aanswer must be <math>\boxed{\textbf{(A) }10}</math>.<br />
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==Video Solution==<br />
https://youtu.be/9k59v-Fr3aE<br />
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==See also==<br />
{{AMC8 box|year=2020|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Mathboy2788