# Difference between revisions of "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. ! LaTeX Error: Bad math environment delimiter.)

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