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

(Created page with '== Problem 5 == 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…')
 
 
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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>?
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== Solution ==
 
== Solution ==
  
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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{(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{\textbf{(E)}\ } 224}.</math>
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==Video Solution==
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https://youtu.be/b3Vorx_bnpU
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~savannahsolver
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== See Also==
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{{AMC10 box|year=2011|ab=B|num-b=4|num-a=6}}
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{{MAA Notice}}

Latest revision as of 12:07, 24 January 2024

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

Video Solution

https://youtu.be/b3Vorx_bnpU

~savannahsolver

See Also

2011 AMC 10B (Problems • Answer Key • Resources)
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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