In Russia you get into a bus, take a ticket, and sometimes say : Wow, a lucky number! Bus tickets are numbered by 6-digit numbers, and a lucky ticket has the sum of 3 first digits being equal to the sum of 3 last digits. When we were in high school we had to write a code that prints out all the lucky tickets' numbers; at least I did, to show my loyalty to the progammers' clan. Now, if you add up all the lucky tickets' numbers you will find out that 13 (the most unlucky number) is a divisor of the result. Can you prove it (without writing a code)?

Let first number of {} = 26 and add 13 to every number after 26 stopping at 52, and eliminate any number not ending in an even integer and ending in 0 to get the sum set. Stop at 998 to get 52 as the sum of each 3-number sets. 52/13= 4. There is just 26,39, and 52 as possibilities. 998+998=52 while 999+999=54 and not divisible by 13. And 54 is the highest number one can whereby add adding single digits in a series of three.

## Comments

Thank you for your answer. Unfortunately, I have still have not gotten what I needed. I am still not aware of how.. if you add all the lucky numbers... the sum is divisible by 13. Do you think you can help me prove that with a little more clarification?

Thanks! :)

I am wondering too since I can't figure why 54 divisible by 17 isn't a possible winninumber red. there seems to be missing criteria. If 13 is necessary, then wheArdis the statement that makes us eliminate 54 as a possible sum to use.

huh?

Im not sure I understand what you're trying to say. :/