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

## 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 $\boxed{\textbf{(D) }99}.$

## See Also

 2014 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 5 Followed byProblem 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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