2014 AMC 12A Problem 15

Attempt Problem 15 of the 2014 AMC 12A below, then check your answer against the professionally curated solution 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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

15.

A five-digit palindrome is a positive integer with respective digits abcba,abcba, where aa is not zero. Let SS be the sum of all five-digit palindromes. What is the sum of the digits of S?S?
99
1818
2727
3636
4545
Answer: B
Concepts:palindromeplace valuesummation
Small hint:
A palindrome abcba‾\overline{abcba} equals 10001a+1010b+100c10001a+1010b+100c
Big hint:
Each aa appears in 10⋅1010\cdot10 palindromes, each bb in 9⋅109\cdot10, each cc in 9⋅109\cdot10; combine with the digit sum 4545
Solution:
Write abcba‾=10001a+1010b+100c.\overline{abcba}=10001a+1010b+100c. Summing over all palindromes, each value of a∈{1,…,9}a\in\{1,\dots,9\} occurs with 10⋅10=10010\cdot10=100 choices of b,c,b,c, and each value of bb or cc occurs with 9⋅10=909\cdot10=90 choices of the other two digits.
Using ∑a=∑b=∑c=45,\sum a=\sum b=\sum c=45, S=45(10001⋅100+1010⋅90+100⋅90)=45⋅1,100,000=49,500,000. \begin{gathered} \scriptsize S=45\big(10001\cdot100+1010\cdot90+100\cdot90\big)\\ =45\cdot1{,}100{,}000\\ =49{,}500{,}000. \end{gathered}
The sum of the digits of SS is 4+9+5=18.4+9+5=18.
Thus, the correct answer is B.
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