2009 AMC 8 Problems/Problem 16

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AJHSME/AMC 8 Problems and Solutions

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