# $109088$ as sum of $3$ palindromes

One can write the number $$109088$$ as the sum of two palindromic numbers as follows: $$109088 = 99199 + 9889$$ Note that the numbers on the right are both palindromes (read the same left to right as right to left), they have different lengths (one is $$5$$ digits long, the other only $$4$$), and both are shorter than the number $$109088$$ itself.

Now try it with three:

Write $$109088$$ as the sum of three palindromic numbers, where the palindromes are of different lengths and shorter than $$6$$ digits.

• I have added the alphametic tag, because it has the feel of one even though it does not explicitly have letters standing in for the digits. I used a computer to find a number with a unique solution, but I'm pretty sure it should be solvable without computer assistance though I haven't really tried it myself. – Jaap Scherphuis Sep 24 '18 at 14:15
• There's a published algorithm for this task, and an (obnoxious) website that implements the algorithm. Doing it computerlessly feels fairly painful to me... – Gareth McCaughan Sep 24 '18 at 14:24
• @GarethMcCaughan: I know, but the fact that the palindromes have to be of different lengths makes it interesting. Without that, there are lots of easy solutions just by splitting the two-palindrome solution into three. – Jaap Scherphuis Sep 24 '18 at 14:27
• Yes, I agree that that makes a difference. And Oray's fairly quick solution suggests that it's less painful than I feared :-). – Gareth McCaughan Sep 24 '18 at 14:31
• For more info about that algorithm Gareth referred to, you can watch this Numberphile video. It was what inspired me to try to make this puzzle based on the idea, but it turned out to be difficult to construct something with a unique solution. – Jaap Scherphuis Sep 24 '18 at 14:57

$$99899+8998+191=109088$$

The reason is

to make to 109k, you have to have at least one 5 digit number with starting 9 and ending 9. otherwise you will not able to reach to 109k and have different type of digits for other numbers.

And

The second number with 4 digits cannot start with $$9$$ because the last one's first and last digit becomes $$0$$ since the result number ends with $$8$$. So it should be different than $$9$$, which could be like $$8$$.

The rest is

Trial and error to be honest.

• Just got this answer but you beat me to it well done! – gabbo1092 Sep 24 '18 at 14:31
• @gabbo1092 :) probably I saw it first, otherwise you would submit it first in my opinion. – Oray Sep 24 '18 at 14:36
• Fun fact: this is the only solution. (I cheated and used a computer) – Luke C. J. Currie Sep 24 '18 at 14:39
• Well done! I guess I should have looked for a longer number with a unique solution :-). – Jaap Scherphuis Sep 24 '18 at 15:02
• For example, I maybe should have asked for $1090288$ as the sum of three palindromes of different lengths, all shorter than 7 digits. – Jaap Scherphuis Sep 24 '18 at 15:49

The following is clearly not what is meant, but a correct solution according to the given rules:

$$109088 = 99199 + 9889 + 0$$

This is a solution because:

• $$0$$ is clearly a palindrome. As are $$99199$$ and $$9889$$, of course.

• The numbers $$99199$$, $$9889$$ and $$0$$ are of different length.

• Each of the three numbers has less than $$6$$ digits.

• No further conditions were given that would rule out this solution.

• Creative? Check. Technically Correct? Check. Amusing? Check. Please, have an upvote (+1). – El-Guest Sep 24 '18 at 16:45
• Fair enough. I should have realised, since this is exactly how I found the two-palindrome solution in the output of my computer program... – Jaap Scherphuis Sep 24 '18 at 20:22