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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![\[+ 91319 + 91419 + 91519 + 91619 + 91719 + 91819 + 91919 + 92029 + 92129 + 92229 + 92329 + 92429 + 92529 + 92629 + 92729 + 92829 + 92929 + 93039 + 93139 + 93239 + 93339 + 93439 + 93539 + 93639 + 93739 + 93839 + 93939 + 94049\]](http://latex.artofproblemsolving.com/4/b/8/4b8fdd4deee360f38fc90b05b3b77b506da1c0d1.png)
![\[+ 94149 + 94249 + 94349 + 94449 + 94549 + 94649 + 94749 + 94849 + 94949 + 95059 + 95159 + 95259 + 95359 + 95459 + 95559 + 95659 + 95759 + 95859 + 95959 + 96069 + 96169 + 96269 + 96369 + 96469 + 96569 + 96669 + 96769 + 96869\]](http://latex.artofproblemsolving.com/7/e/6/7e6b31bebac0370a0e910a40ae74d165c58af755.png)
![\[+ 96969 + 97079 + 97179 + 97279 + 97379 + 97479 + 97579 + 97679 + 97779 + 97879 + 97979 + 98089 + 98189 + 98289 + 98389 + 98489 + 98589 + 98689 + 98789 + 98889 + 98989 + 99099 + 99199 + 99299 + 99399 + 99499 + 99599 + 99699\]](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 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
| 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 | |
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