18 Jan '05 15:33

Suppose you have all the numbers from 0 to (2^n) - 1 (n>=2) and you want to divide them into two even groups so that:

The sum of the m-th powers of the numbers in the first group equals the sum of the m-th power of the numbers in the second group, for all natural m<=n-1. /That's a1^m+a2^m+...+(a2^(n-1))^m = b1^m + b2^m+...+(b2^(n-1))^m) if a1,a2,.., a2^(n-1) belong to the first group, and b1,b2, b3,.., b2^(n-1) belong the second/.

For example:

1)2^2 -1 =3, and the groups are 0,3 and 1,2. /1+2=3+0/

2) 2^3 -1 =7, and the groups are 0,3,5,6 and 1,2,47

/0+3+5+6 = 1+2+4+7 = 14; 0^2+3^2+5^2 + 6^2 = 70 = 1^2 + 2^2 + 4^2 + 7^2 /

There really is a simple way to find what numbers go to the first and to the second group. Can you find it ?

The sum of the m-th powers of the numbers in the first group equals the sum of the m-th power of the numbers in the second group, for all natural m<=n-1. /That's a1^m+a2^m+...+(a2^(n-1))^m = b1^m + b2^m+...+(b2^(n-1))^m) if a1,a2,.., a2^(n-1) belong to the first group, and b1,b2, b3,.., b2^(n-1) belong the second/.

For example:

1)2^2 -1 =3, and the groups are 0,3 and 1,2. /1+2=3+0/

2) 2^3 -1 =7, and the groups are 0,3,5,6 and 1,2,47

/0+3+5+6 = 1+2+4+7 = 14; 0^2+3^2+5^2 + 6^2 = 70 = 1^2 + 2^2 + 4^2 + 7^2 /

There really is a simple way to find what numbers go to the first and to the second group. Can you find it ?