- 15 May '05 07:54The set (1,2,3,4) can be partioned into two subsets (1,4) and (2,3) of the same size. 1+4=2+3

a)find the next whole number n, above 4, for which the set (1,2,...n) can be paritioned into two subsets S and T of the same size, with the sum of S and T the same.

b)find all partition 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.

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

- the sum of the numbers in S equals the sum of T

-the sum of the squares of the number in S equals the sum of the squares of the numbers in T.

-the sum of the cubes of the numbers in S equals the sum of the cubes of the numbers in T.

c) show terri is right.

Willy says she can partition the set (1,2...8) into subsets S and T, not necessarily the same size so that:

- the sum of the numbers in S equals the sum of T

-the sum of the squares of the number in S equals the sum of the squares of the numbers in T.

-the sum of the cubes of the numbers in S equals the sum of the cubes of the numbers in T.

d)show why you dont belive willy - 15 May '05 08:52

It appears as though you and phgao are both trying to cheat on the same 2005 Intermediate Maths Challenge, whatever that is. I am just basing this on 12122000's post in phgao "sets" thread, but i do find it quite interesting that your question matches his more or less exactly.*Originally posted by darkmage***The set (1,2,3,4) can be partioned into two subsets (1,4) and (2,3) of the same size. 1+4=2+3**

a)find the next whole number n, above 4, for which the set (1,2,...n) can be paritioned into two subsets S and T of the same size, with the sum of S and T the same.

b)find all partition in a) with the additional property that the sum of the squares of the numbe ...[text shortened]... umbers in S equals the sum of the cubes of the numbers in T.

d)show why you dont belive willy - 16 May '05 01:58

I can solve part "a". You can take any set of consecutive numbers that has a cardinality divisible by 4. So, the next set would be "1 2 3 4 5 6 7 8". Notice that if you reverse the numbers, and add 1 with 8, 2 with 7, 3 with 6, and 4 with 5, you get 9 for each addition. Now, you know your pairs. Because you have 4 subsets of 2, you can group any two subsets of 2 you want, and call them S, and the other two subsets are T. So, you have set S = (1, 8, 2, 7), and T = (3, 6, 4, 5). If you had an original set of twelve, you would use the same trick of reversing the order, and then group 3 subsets of two, to get S, and T would be the rest.*Originally posted by darkmage***The set (1,2,3,4) can be partioned into two subsets (1,4) and (2,3) of the same size. 1+4=2+3**

a)find the next whole number n, above 4, for which the set (1,2,...n) can be paritioned into two subsets S and T of the same size, with the sum of S and T the same.

b)find all partition in a) with the additional property that the sum of the squares of the numbe ...[text shortened]... umbers in S equals the sum of the cubes of the numbers in T.

d)show why you dont belive willy