Lucky Tickets

[BMAD]

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)?

[Prime]

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.

Let's call the first 3 digits of a 6-digit number

h and the last 3 digts t. Then the 6-digit number is 1000*h + t.

Let's construct another number by swapping heads and tails, that is putting t up front and h in the back. Thus the swapped number is 1000*t + h. Clearly, if the ht is a “lucky number” then so is th. Let's add them together:

1000h + t + 1000t + h = 1001(h+t).

Each 6-digit number has exactly one corresponding swapped number. Some numbers equal to their swapped number. Example: 532,532. Such symmetrical lucky numbers are obtained by multiplying any 3-digit number by 1001: 532*1001 = 532,532.

Thus the set all lucky numbers may be presented as a set of all symmetrical numbers, which are divisible by 1001 and a set of pairs of swapped numbers, where the sum of each pair is also divisible by 1001.

1001 = 7*11*13.

Q.E.D.

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/forum/index.php/topic/15583-more-divisibility/?p=328196]More divisibility

Edited March 10, 2013 by Prime

[Prime]

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?

Will you also find that the sum is divisible by 7 and 11?

[bonanova]

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

[BMAD]

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.

Let's call the first 3 digits of a 6-digit number

h and the last 3 digts t. Then the 6-digit number is 1000*h + t.

Let's construct another number by swapping heads and tails, that is putting t up front and h in the back. Thus the swapped number is 1000*t + h. Clearly, if the ht is a lucky number then so is th. Let's add them together:

1000h + t + 1000t + h = 1001(h+t).

Each 6-digit number has exactly one corresponding swapped number. Some numbers equal to their swapped number. Example: 532,532. Such symmetrical lucky numbers are obtained by multiplying any 3-digit number by 1001: 532*1001 = 532,532.

Thus the set all lucky numbers may be presented as a set of all symmetrical numbers, which are divisible by 1001 and a set of pairs of swapped numbers, where the sum of each pair is also divisible by 1001.

1001 = 7*11*13.

Q.E.D.

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/forum/index.php/topic/15583-more-divisibility/?p=328196]More divisibility

Perm Russia actually

Edited March 11, 2013 by BMAD