2014 AMC 12A Problems/Problem 15

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 1

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 $\boxed{\textbf{(B)}\; 18}$ .

Solution 2

Notice that $10001+ 99999 = 110000.$ In fact, ordering the palindromes in ascending order, we find that the sum of the nth palindrome and the nth to last palindrome is $110000.$ We have $9*10*10$ palindromes, or $450$ pairs of palindromes summing to $110000.$ Performing the multiplication gives $49500000$ , so the sum $\boxed{\textbf{(B)}\; 18}$ .

Solution 3

As shown above, there are a total of $900$ five-digit palindromes. We can calculate their sum by finding the expected value of a randomly selected palindrome satisfying the conditions given, then multiplying it by $900$ to get our sum. The expected value for the ten-thousands and the units digit is $\frac{1+2+3+\cdots+9}{9}=5$ , and the expected value for the thousands, hundreds, and tens digit is $\frac{0+1+2+\cdots+9}{10}=4.5$ . Therefore our expected value is $5\times10^4+4.5\times10^3+4.5\times10^2+4.5\times10^1+5\times10^0=55,\!000$ . Since the question asks for the sum of the digits of the resulting sum, we do not need to keep the trailing zeros of either $55,\!000$ or $900$ . Thus we only need to calculate $55\times9=495$ , and the desired sum is $\boxed{\textbf{(B) }18}$ .

Solution 4 (Variation of #2)

First, allow $a$ to be zero, and then subtract by how much we overcount. We'll also sum each palindrome with its $\textit{complement}$ . If $abcba$ is a palindrome, then its complement is $defed$ where $d=9-a$ , $e=9-b$ , $f=9-c$ . Notice how every palindrome has a unique compliment, and that the sum of a palindrome and its complement is $99999$ . Therefore, the sum of our palindromes is $99999\times (10^3/2)$ . (There are $10^3/2$ pairs.)

However, we have overcounted, as something like $05350$ $\textit{isn't}$ a palindrome by the problem's definition, but we've still included it. So we must subtract the sum of numbers in the form $nmn0$ . By the same argument as before, these sum to $9990\times (10^2/2)$ . Therefore, the sum that the problem asks for is:

$500\times99999-50\times 9990$ $=500\times99999-500\times 999$ $=500(99999-999)$ $=500\times 99000$

Since all we care about is the sum of the digits, we can drop the $0$ 's.

$5\times99$ $=5\times(100-1)$ $=495$

And finally, $4+9+5=\boxed{\textbf{(B)}18}$

2014 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

Page

Toolbox

Search