- 06 May '05 10:45The set (1,2,3,4) can be partitioned into two subsets (1,4) and (2,3) of the same size. Note that 1 + 4 = 2 + 3

Qa. Find the next whole number n, above 4, for which the set (1,2,...,n) can be partitioned into two subsets S and T of the same size, with the sum of the numbers in S equal to the sum of the numbers in T. (this is farily simple as n can't be odd, and the sum of 1,...n can't be odd either)

Qb. Find all partitions in a with the ADDITIONAL property that the sum of the squares of the numbers in S equals the sum of the squares of the numbers in T.

Jonny says that he can partition the set (1,2,...,16) into two subsets S and T of the same size so that:

A: the sum of the numbers in S equals the sum of the numbers in T

B: the sum of the squares of the numbers in S equals the sum of the squares in T

c: the sum of the cubes of the numbers in S equals the sum of the cubes in T

Qc. Show/prove that Jonny is correct.

He then says he can partition the set (1,2,...,8) into two subsets S and T, not necessarily of the same size, so that:

A: the sum of the numbers in S equals the sum of the numbers in T

B: the sum of the squares of the numbers in S equals the sum of the squares in T

c: the sum of the cubes of the numbers in S equals the sum of the cubes in T

Qd: Explain why you do not believe him this time. - 09 May '05 09:02

"phgao", it intrigues me that some of your problems match those given out in the intermediate paper of the 2005 maths challenge stage, shortly after it has been released. You are aware that anyone entering in this is not allowed to discuss the answer with another human, only questions 5 and 6 may be discussed with one other person who is participating in the challenge. This is goes also for the "noether" student problems, except answers may not be discussed. I hope you understand that if this continues we will have to inform the authorities.*Originally posted by phgao***The set (1,2,3,4) can be partitioned into two subsets (1,4) and (2,3) of the same size. Note that 1 + 4 = 2 + 3**TT

Qa. Find the next whole number n, above 4, for which the set (1,2,...,n) can be partitioned into two subsets S and T of the same size, with the sum of the numbers in S equal to the sum of the numbers in T. (this is farily simple as n can't be odd ...[text shortened]... ers in S equals the sum of the cubes in T

Qd: Explain why you do not believe him this time.

p.s. Changing names from the problems still counts as CHEATING. If you have recieved this message and are willing to stop CHEATING, please reply, so that no further steps will be taken

Thankyou for your cooperation.