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

Revision as of 16:11, 4 June 2011

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

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)
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All AMC 10 Problems and Solutions

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−	== Problem 5 ==	+	== Problem==

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

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

Revision as of 16:11, 4 June 2011

Problem

Solution

See Also