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 $b$ . $\longrightarrow 7 \times 32 = \boxed{\mathrm{(E) \ } 224}$

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

Problem 5

Solution

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