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Lucky Tickets


Best Answer Prime, 11 March 2013 - 12:24 AM

Going with the following assumptions as answers to my questions:
1. Leading zeros are allowed.
2. We work with the sum of numbers, not digits.
3. The school No. 7 is in Moscow in the vicinity of MSU (МГУ) and Vernadski Prospect.
 

Spoiler for proof

 
I have a different proof of the same, which gives even more insight into divisibility rules, but it would take longer to explain.
We have another Divisibility puzzle going here: http://brainden.com/...re divisibility Go to the full post


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#1 BMAD

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Posted 10 March 2013 - 04:21 AM

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 (guys from math school No. 7 might remember that ) 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 Prime

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Posted 10 March 2013 - 07:51 PM

Do tickets include numbers with leading zeros?

Do you add together all actual 6-digit numbers, or just the digits from those numbers?

In which town math school No. 7 is located?

 

Spoiler for in addition

 


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Past prime, actually.


#3 bonanova

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Posted 10 March 2013 - 11:33 PM

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 (guys from math school No. 7 might remember that ) 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)? 

 

If I understand, a ticket has the numerical string 'abcdef', where

  1. a-f are in the set { 0 1 2 3 4 5 6 7 8 9 },
  2. a+b+c = d+e+f
  3. duplicates are allowed.

To re-state Prime's question, I will ask: is 000000 a lucky ticket?

 

When you say "all the lucky ticket numbers" sum to a multiple of 13,

which do you  mean?

  1. for each and every lucky ticket abcdef, a+b+c+d+e+f is divisible by 13, or
  2. taking all possible lucky numbers, their sum, abcdef + ghijkl + ... + uvwxyz is divisible by 13

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#4 Prime

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Posted 11 March 2013 - 12:24 AM   Best Answer

Going with the following assumptions as answers to my questions:
1. Leading zeros are allowed.
2. We work with the sum of numbers, not digits.
3. The school No. 7 is in Moscow in the vicinity of MSU (МГУ) and Vernadski Prospect.
 
Spoiler for proof

 
I have a different proof of the same, which gives even more insight into divisibility rules, but it would take longer to explain.
We have another Divisibility puzzle going here: http://brainden.com/...re divisibility

Edited by Prime, 11 March 2013 - 12:31 AM.

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Past prime, actually.


#5 BMAD

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Posted 11 March 2013 - 02:46 AM

Going with the following assumptions as answers to my questions:
1. Leading zeros are allowed.
2. We work with the sum of numbers, not digits.
3. The school No. 7 is in Moscow in the vicinity of MSU (МГУ) and Vernadski Prospect.

Perm Russia actually.
 

Spoiler for proof

 
I have a different proof of the same, which gives even more insight into divisibility rules, but it would take longer to explain.
We have another Divisibility puzzle going here: http://brainden.com/...re divisibility


Perm Russia actually

Edited by BMAD, 11 March 2013 - 02:47 AM.

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