Lucky Lottery Tickets

Six-digit lottery ticket numbers are considered lucky if the sum of the first three numbers is equal to the last three. Eg. 123204, 871529...
Prove that all possible lucky numbers added together is divisible by thirteen.

• Can the number start with zeroes? Oct 19, 2017 at 20:09
• I think so, that's what I assumed
– user35295
Oct 19, 2017 at 20:13
• Warning - lucky numbers are a different thing in mathematics. Oct 19, 2017 at 20:24
• Has a correct answer been given? If so, please don't forget to $\color{green}{\checkmark \small\text{Accept}}$ it :)
– Rubio
Oct 23, 2017 at 2:53

If $abcxyz$ is a lucky number (where each letter represents a digit), then so is $xyzabc$. So most lucky numbers occur in pairs (e.g. $123204$ and $204123$), but there are also some that pair up with themselves because they have two equal halves (e.g. $123123$). The latter is divisible by $1001$ and hence by $13$ as well ($abcabc = 1001*abc = 13*77*abc$). The lucky pairs together are also divisible by $13$ (because $abcxyz+xyzabc = abcabc+xyzxyz = 13*77*(abc+xyz)$).

Therefore the sum of all lucky numbers is also divisible by $13$.

Note that this assumes that lucky numbersare allowed to have leading zeroes. For example, $123042$ forms a lucky pair with $042123$. If no leading zeroes are allowed then not all lucky numbers form a pair and the sum of the valid ones will probably not be divisible by $13$.

• ^vote with a note: Excellent point about leading 0s, Jaap S. Now that you mention them, they do seem to stretch the colloquial meaning of 6-digit numbers.
– humn
Oct 19, 2017 at 20:29
• Are two or more leading zeros eligible for such lucky numbers? Oct 20, 2017 at 5:06
• Yes, leading zeroes are allowed. Oct 27, 2017 at 14:44

It is divisible by 1001, therefore with 13 as well(1001=7*11*13). If the sum(which is the same) is x, there are ways to do so. For example: x=25:9+9+7, 9+8+8, which is used in all permutation in first and last numbers, equal times, which means the sum of these 6digit numbers is divisible by 1001.

• Yes, but adding it with all other x=6 ones the sum will be. 123 in the first digit is happening the same times as in the last digits. 204 likewise. Oct 19, 2017 at 14:34
• AAhhhh...never mind... I misunderstood the question Oct 19, 2017 at 14:36
• Not all lucky numbers are divisible by 1001. Oct 19, 2017 at 15:58
• @Jacob That's not what the answer is saying. It's saying that rnpu crezhgngvba bs guerr ahzoref vf hfrq na rdhny nzbhag bs gvzrf va gur svefg guerr ahzoref nf va gur ynfg guerr ahzoref. Gurersber, gur fhz bs nyy fhpu ahzoref vf qvivfvoyr ol 1001. (rot13)
– Apep
Oct 19, 2017 at 20:05
• There might be a valid idea behind this answer, but the wording is incomprehensible. Oct 19, 2017 at 20:51

Since 13 is not divisible by 2, both sides of the lucky number don't matter. We consider the front side. Consider how different ways to arrange the digits. 000, 100, 010, 001 for example or you can consider the other end of the spectrum; 999, 998, 989, 899.

Pairing these up, you will get pairs of 999 and one 999. Notice that 999 999 is divisible by 13.

Consider how many different ways there are to create pairs. We can use the multiplicity rule.

First digit: 10 ways Second: 10 ways Third: 10 ways, thus, there are 1000 ways to create these pairs without repetition.

But, remember, we still have 999.

Thus the sum is 999*1000 = 999 000 Don't forget! We still have 999 so the real sum of the first 3 digits of all lucky numbers is 999 999.

Remember what I said earlier?

Notice that 999 999 is divisible by 13.

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Therefore, the sum of all lucky numbers is divisible by 13.

But wait you say: You forgot about the other side!

Well, actually, the sum would be the same.

But wait you say again: You forgot that they are in the Thousandth place!

Well, actually, that would just be 999 999 * 1000 is divisible because 999 999 is divisible.