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Difference between revisions of "2009 AMC 8 Problems/Problem 16"

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How many <math> 3</math>-digit positive integers have digits whose product equals <math> 24</math>?
 
How many <math> 3</math>-digit positive integers have digits whose product equals <math> 24</math>?
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<math> \textbf{(A)}\ 12 \qquad \textbf{(B)}\ 15 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ 21 \qquad \textbf{(E)}\ 24</math>
 
<math> \textbf{(A)}\ 12 \qquad \textbf{(B)}\ 15 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ 21 \qquad \textbf{(E)}\ 24</math>
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==Solution==
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With the digits listed from least to greatest, the <math>3</math>-digit integers are <math>138,146,226,234</math>. <math>226</math> can be arranged in <math>\frac{3!}{2!} = 3</math> ways, and the other three can be arranged in <math>3!=6</math> ways. There are <math>3+6(3) = \boxed{\textbf{(D)}\ 21}</math> <math>3</math>-digit positive integers.
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==See Also==
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{{AMC8 box|year=2009|num-b=15|num-a=17}}
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{{MAA Notice}}

Latest revision as of 20:22, 28 August 2016

Problem

How many $3$-digit positive integers have digits whose product equals $24$?


$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 15 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ 21 \qquad \textbf{(E)}\ 24$

Solution

With the digits listed from least to greatest, the $3$-digit integers are $138,146,226,234$. $226$ can be arranged in $\frac{3!}{2!} = 3$ ways, and the other three can be arranged in $3!=6$ ways. There are $3+6(3) = \boxed{\textbf{(D)}\ 21}$ $3$-digit positive integers.

See Also

2009 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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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