2014 AMC 12A Problem 6

Below is the professionally curated solution for Problem 6 of the 2014 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2014 AMC 12A solutions, or check the answer key.

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Concepts:place value digits

Difficulty rating: 1270

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?

$44$

$55$

$77$

$99$

$110$

Solution:

Let the larger number be $10a+b.$ Then $(10a+b)-(10b+a)=9(a-b)=5(a+b),$ which simplifies to $2a=7b.$

The only nonzero digits satisfying this are $a=7$ and $b=2,$ so the number is $72$ and its reverse is $27.$

Their sum is $72+27=99.$

Thus, the correct answer is D.

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