Difference between revisions of "2014 AMC 12A Problems/Problem 6"

 
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==Solution 2 (Meta)==
 
==Solution 2 (Meta)==
 
We start like above. Let the two digits be <math>a</math> and <math>b</math>. Therefore, <math>5(a+b) = 10a+b-10b-a=9(a-b)</math>. Since we are looking for <math>10a+b+10b+a=11(a+b)</math> and we know that <math>a+b</math> must be a multiple of <math>9</math>, the only answer choice that works is <math>\boxed{\textbf{(D) }99}.</math>
 
We start like above. Let the two digits be <math>a</math> and <math>b</math>. Therefore, <math>5(a+b) = 10a+b-10b-a=9(a-b)</math>. Since we are looking for <math>10a+b+10b+a=11(a+b)</math> and we know that <math>a+b</math> must be a multiple of <math>9</math>, the only answer choice that works is <math>\boxed{\textbf{(D) }99}.</math>
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==Video Solution==
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https://youtu.be/rJytKoJzNBY
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2014|ab=A|num-b=5|num-a=7}}
 
{{AMC12 box|year=2014|ab=A|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 23:15, 25 November 2020

Problem

The difference between a two-digit number and the number obtained by reversing its digits is $5$ times the sum of the digits of either number. What is the sum of the two digit number and its reverse?

$\textbf{(A) }44\qquad \textbf{(B) }55\qquad \textbf{(C) }77\qquad \textbf{(D) }99\qquad \textbf{(E) }110$

Solution 1

Let the two digits be $a$ and $b$. Then, $5a + 5b = 10a + b - 10b - a = 9a - 9b$, or $2a = 7b$. This yields $a = 7$ and $b = 2$ because $a, b < 10$. Then, $72 + 27 = \boxed{\textbf{(D) }99}.$

Solution 2 (Meta)

We start like above. Let the two digits be $a$ and $b$. Therefore, $5(a+b) = 10a+b-10b-a=9(a-b)$. Since we are looking for $10a+b+10b+a=11(a+b)$ and we know that $a+b$ must be a multiple of $9$, the only answer choice that works is $\boxed{\textbf{(D) }99}.$

Video Solution

https://youtu.be/rJytKoJzNBY

See Also

2014 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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 12 Problems and Solutions

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