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 ![\[(1+2+\cdots+9)(100)(10^4+1)+(1+2+\cdots+9)(90)(10^3+10)+(1+2+\cdots+9)(90)(10^2)=49500000\]](http://latex.artofproblemsolving.com/3/8/b/38b77fb7c65e53c50aeee86e0951039d75743e95.png)
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.
![\[10001 + 10101 + 10201 + 10301 + 10401 + 10501 + 10601 + 10701 + 10801 + 10901 + 11011 + 11111 + 11211 + 11311 + 11411 + 11511 + 11611 + 11711 + 11811 + 11911 + 12021 + 12121 + 12221 + 12321 + 12421 + 12521 + 12621 + 12721 + 12821\]](http://latex.artofproblemsolving.com/0/4/e/04ed57269d3251a56c2000f49ef6a5f5e3b5fdf8.png)
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![\[+ 96969 + 97079 + 97179 + 97279 + 97379 + 97479 + 97579 + 97679 + 97779 + 97879 + 97979 + 98089 + 98189 + 98289 + 98389 + 98489 + 98589 + 98689 + 98789 + 98889 + 98989 + 99099 + 99199 + 99299 + 99399 + 99499 + 99599 + 99699\]](http://latex.artofproblemsolving.com/a/0/c/a0c0df2e53a22ba37308bc453fcf3f8c4bfebcae.png)
![\[+ 99799 + 99899 + 99999 = 49500000 = 4 + 9 + 5 = 18 \to \boxed{(B)}\]](http://latex.artofproblemsolving.com/5/0/c/50cba46f9ce5481d6d4eac0978cc8392688b7646.png)
.
Solution Three
Notice that 
In fact, ordering the palindromes in ascending order, we find that the sum of the nth palindrome and the nth to last palindrome is 
We have 
palindromes, or 
pairs of palindromes summing to Performing the multiplication gives 
, so the sum is .
See Also
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
