2014 AMC 12A Problem 15

Below is the professionally curated solution for Problem 15 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:palindrome place value summation

Difficulty rating: 1660

15.

A five-digit palindrome is a positive integer with respective digits $abcba,$ where $a$ is not zero. Let $S$ be the sum of all five-digit palindromes. What is the sum of the digits of $S?$

$9$

$18$

$27$

$36$

$45$

Solution:

Write $\overline{abcba}=10001a+1010b+100c.$ Summing over all palindromes, each value of $a\in\{1,\dots,9\}$ occurs with $10\cdot10=100$ choices of $b,c,$ and each value of $b$ or $c$ occurs with $9\cdot10=90$ choices of the other two digits.

Using $\sum a=\sum b=\sum c=45,$ $S=45\big(10001\cdot100+1010\cdot90+100\cdot90\big)=45\cdot1{,}100{,}000=49{,}500{,}000.$

The sum of the digits of $S$ is $4+9+5=18.$

Thus, the correct answer is B.

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