2014 AMC 12A Problems/Problem 15

Revision as of 11:12, 2 February 2021 by Captainsnake (talk | contribs) (Added solution)

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

See Also

2014 AMC 12A (ProblemsAnswer KeyResources)
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

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