## Monday, March 12, 2007

### Sums, Squares and Somethings

Partition {1,2,..,16} into 2 sets such that the sum of the numbers in one set equals that in the other, and likewise for sum of squares and sum of cubes.

Anonymous said...

A={1,4,6,7,10,11,13,16}
B={2,3,5,8,9,12,14,15}
should do it.

8:55 AM
Anonymous said...

For Squares A={16,15,13,9,4,1} B={2,3,5,6,7,8,10,11,12,14}

Left the cubes case for the third person.

-- second person

12:51 PM
Anonymous said...

First Poster:

Brilliant solution. I misunderstood the problem statement.

-- Second poster/person

1:24 PM
metoo said...

Consider the set {1,....,n} and partition it into 2 sets so that the sum of i-powers of the numbers in the two sets are the same, for i=0,1,2, ....

9:42 PM
Anonymous said...

metoo, my guess is that you can only consider the first ~log(n)-1 moments; in which case it does not seem to be hard to find a solution (simple generalization of the case n = 2^k: the solution will be inductively constructed - express n as 2^k(2q-1) for the generic even case). Do you have a prove of uniqueness - even for the perfect power of 2 case?

2:37 PM
Anonymous said...

sorry, I meant "Do you have a prove of uniqueness for the perfect power of 2 case?" Thanks :)

2:39 PM