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Higher Level Math Question---> Russian Lucky Numbers?!

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)?

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2 Answers

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. 

Comment

Hello Miss.,

This question from nearly a week ago seems never to have been resolved favorably, so I will attempt to give a brief exposition of its solution. 

First, we must be clear on what constitutes a bus ticket.  From the wording of the problem, it seems to me that any string of six digits, from 000000 to 999999 is a valid ticket, since this would enable the bus system to have an ordering of tickets.  In other words, tickets beginning with one or more zeroes are perfectly acceptable.

If we agree to this definition of a ticket, the problem becomes straightforward.  Let an arbitrary lucky bus ticket be denoted by its six-digits as abcdef.  Write the value of the ticket as follows:

1000abc + def

where we think of abc and def as normal three-digit integers.  Now, if abcdef is lucky, then so is defabc.  Note that if we add together the values of the two lucky tickets abcdef and defabc, we get

1001(abc + def),

and since 1001 is divisible by 13, then any pair of lucky tickets of this type sum to a multiple of 13.  The last thing to note is that any ticket of the form abcabc will only occur once in the sum of lucky tickets, not as a pair of tickets.  However, tickets of this form are already divisible by 13, since they have the value

1000abc + abc = 1001abc.

So, the sum of all the lucky bus tickets must be divisible by 13. 

Since 1001 = 7·11·13, the sum of all the bus tickets is also divisible by 7 and 11, we could say that since 7 and 11 are (maybe) lucky numbers, these outweigh the unlucky influence of 13!

I hope this explanation is clear.

Regards,

Hassan H.