2014 AMC 12A Problems/Problem 15

Problem

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

Solution One

For each digit there are (ways of choosing and ) palindromes. So the s contribute to the sum. For each digit there are (since ) palindromes. So the s contribute to the sum. Similarly, for each there are palindromes, so the contributes to the sum.

It just so happens that so the sum of the digits of the sum is , or .

(Solution by AwesomeToad)

Solution Two

As there are only five digit palindromes, it is sufficient to add up all of them.

\[+ 99799 + 99899 + 99999 = 49500000 = 4 + 9 + 5 = \Boxed{18} \to B\] (Error compiling LaTeX. Unknown error_msg)

.

Solution Three

Notice that 10001+ 99999 = 111000. In fact, ordering the palindromes in ascending order, we find that the sum of the nth palindrome and the nth to last palindrome is 111000. We have 9*10*10 palindromes, or 450 pairs of palindromes summing to 111000. Performing the multiplication gives 49500000, so the sum is 18.

See Also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.