# Difference between revisions of "2011 AMC 10B Problems/Problem 5"

## Problem 5

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

Since the $161$ was the erroneous product of two integers, one that had two digits, the two factors have to be $7$ and $23$. Reverse the digits of the two-digit number so that $a$ is $32$. Find the product of $a$ and $b$. $\longrightarrow 7 \times 32 = \boxed{\mathrm{(E) \ } 224}$