2011 AMC 10B Problems/Problem 5

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

We have $161 = 7 \cdot 23.$ Since $a$ has two digits, the factors must be $23$ and $7,$ so $a = 32$ and $b = 7.$ Then, $ab = 7 \times 32 = \boxed{\mathrm{\textbf{(E)}\ } 224}.$

Video Solution

https://youtu.be/b3Vorx_bnpU

~savannahsolver

2011 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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2011 AMC 10B Problems/Problem 5

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