2011 AMC 12B Problems/Problem 4

Problem

In multiplying two positive integers $a$ and $b$ , Ron reversed the digits of the two-digit number $a$ . His erroneous product was $161.$ What is the correct value of the product of $a$ and $b$ ?

$\textbf{(A)}\ 116 \qquad \textbf{(B)}\ 161 \qquad \textbf{(C)}\ 204 \qquad \textbf{(D)}\ 214 \qquad \textbf{(E)}\ 224$

Solution

Taking the prime factorization of $161$ reveals that it is equal to $23*7.$ Therefore, the only ways to represent $161$ as a product of two positive integers is $161*1$ and $23*7.$ Because neither $161$ nor $1$ is a two-digit number, we know that $a$ and $b$ are $23$ and $7.$ Because $23$ is a two-digit number, we know that a, with its two digits reversed, gives $23.$ Therefore, $a = 32$ and $b = 7.$ Multiplying our two correct values of $a$ and $b$ yields

$a*b = 32*7 =$

\[= \boxed{224\  \(\textbf{(E)}}\] (Error compiling LaTeX. Unknown error_msg)

2011 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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All AMC 12 Problems and Solutions

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2011 AMC 12B Problems/Problem 4

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