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{\mathrm{(E) \ } 224}</math>
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We have <math>161 = 7 \cdot 23.</math> Since <math>a</math> has two digits, the factors must be <math>23</math> and <math>7,</math> so <math>a = 32</math> and <math>b = 7.</math> Then, <math>ab = 7 \times 32 = \boxed{\mathrm{(E) \ } 224}.</math>
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==Video Solution==
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https://youtu.be/b3Vorx_bnpU
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~savannahsolver
  
 
== See Also==
 
== See Also==

Latest revision as of 12:29, 7 November 2020

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, $ab = 7 \times 32 = \boxed{\mathrm{(E) \ } 224}.$

Video Solution

https://youtu.be/b3Vorx_bnpU

~savannahsolver

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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