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

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== Solution ==
 
== Solution ==
  
Since the <math>161</math> was the erroneous product of two integers, one that had two digits, the two factors have to be <math>7</math> and <math>23</math>. Reverse the digits of the two-digit number so that <math>a</math> is <math>32</math>. Find the product of <math>a</math> and <math>b</math>. <math>\longrightarrow 7 \times 32 = \boxed{(E) 224}</math>
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Since the <math>161</math> was the erroneous product of two integers, one that had two digits, the two factors have to be <math>7</math> and <math>23</math>. Reverse the digits of the two-digit number so that <math>a</math> is <math>32</math>. Find the product of <math>a</math> and <math>b</math>. <math>\longrightarrow 7 \times 32 = \boxed{\mathrm{(E) \ } 224}</math>
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== See Also==
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{{AMC10 box|year=2011|ab=B|num-b=4|num-a=6}}

Revision as of 16:43, 4 June 2011

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

See Also

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