October 16-22, 2007

The problem:

Suppose raffle tickets are identified by a six-digit number. A ticket (number) is lucky if the sum of the first three digits is equal to the sum of the last three digits. For example, the number 254119 is lucky since 2 + 5 + 4 = 1 + 1 + 9. Show that if you add up all the possible lucky ticket numbers, the sum is divisible by the unlucky number 13.

Solution (from Adam Estrup):

Each lucky number x has an associated lucky number y=999999-x. It's easy to see that y is lucky if and only if x is lucky. It's also easy to see that x and y are distinct with x+y=999999. So, lucky numbers come in pairs, each pair summing to 999999. Let n be the number of such pairs. Then the sum of all possible lucky ticket numbers is 999999n=13(76923)n.

List of Solvers:

Aaron Dilley, Bob Sterling, Michelle Kim (undergraduate); Field Cady, Dustin Moody, Robert Bradshaw, Adam Estrup, Peizhe Shi, Aneesh Hariharan (graduate); Gary Raymond, Marina Meila (faculty); Rich Bauer (alum)