2014 AMC 12A Problems/Problem 15

Revision as of 20:49, 9 February 2014 by Dragon6point1 (talk | contribs) (Solution Two)

Problem

A five-digit palindrome is a positive integer with respective digits $abcba$, where $a$ is non-zero. Let $S$ be the sum of all five-digit palindromes. What is the sum of the digits of $S$?

$\textbf{(A) }9\qquad \textbf{(B) }18\qquad \textbf{(C) }27\qquad \textbf{(D) }36\qquad \textbf{(E) }45\qquad$

Solution One

For each digit $a=1,2,\ldots,9$ there are $10\cdot10$ (ways of choosing $b$ and $c$) palindromes. So the $a$s contribute $(1+2+\cdots+9)(100)(10^4+1)$ to the sum. For each digit $b=0,1,2,\ldots,9$ there are $9\cdot10$ (since $a \neq 0$) palindromes. So the $b$s contribute $(0+1+2+\cdots+9)(90)(10^3+10)$ to the sum. Similarly, for each $c=0,1,2,\ldots,9$ there are $9\cdot10$ palindromes, so the $c$ contributes $(0+1+2+\cdots+9)(90)(10^2)$ to the sum.

It just so happens that \[(1+2+\cdots+9)(100)(10^4+1)+(1+2+\cdots+9)(90)(10^3+10)+(1+2+\cdots+9)(90)(10^2)=49500000\] so the sum of the digits of the sum is $18$, or $\boxed{\textbf{(B)}}$.

(Solution by AwesomeToad)

Solution Two

As there are only $9\cdot10\cdot10 = 900$ five digit palindromes, it is sufficient to add up all of them. $a$.

See Also

2014 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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All AMC 12 Problems and Solutions

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