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

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== Problem 5 ==
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== 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>?

Revision as of 17:11, 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 $b$. $\longrightarrow 7 \times 32 = \boxed{\mathrm{(E) \ } 224}$

See Also

2011 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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 10 Problems and Solutions
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