2010 AMC 10A Problem 9

Below is the professionally curated solution for Problem 9 of the 2010 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2010 AMC 10A solutions, or check the answer key.

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Concepts:palindrome digits bounding to limit cases

Difficulty rating: 1280

A palindrome, such as $83438,$ is a number that remains the same when its digits are reversed. The numbers $x$ and $x + 32$ are three-digit and four-digit palindromes, respectively. What is the sum of the digits of $x?$

$20$

$21$

$22$

$23$

$24$

Solution:

Note that $x$ is at most $999.$ This means that $x + 32$ has a maximum of $1031.$

Similarly, we have that the minimum value of $x + 32$ is $1000.$

The only palindrome in this range is $1001,$ so this is what $x + 32$ equals.

Then $x + 32 = 1001$ $x = 969.$

The sum of the digits is then $9 + 6 + 9 = 24.$

Thus, E is the correct answer.

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