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Posers and Puzzles

Posers and Puzzles

  1. 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 ?
  2. Standard member Palynka
    Upward Spiral
    18 Jan '05 16:03
    well if the numbers are to the power of 2 then we can always group them in subgroups with 4 numbers in sequence.

    If n>=2 then 2^n is divisible by four (2^(n-2) will be an integer). Since we remove n but add 0 it will always be possible to form x subgroups of 4 elements.

    The first group would be (0,1,2,3)
    The second (4,5,6,7)
    Third (8,9.10,11)

    Until (n-1,n-2,n-3,n-4)

    We can now assing to group A the first and last elements of each subgroup and to B the elements in the middle of each subgroup ensuring the sums will be identical.

    PS: I'm sure that someone can explain this more clearly than I did...
  3. 18 Jan '05 16:38 / 1 edit
    Originally posted by Palynka
    well if the numbers are to the power of 2 then we can always group them in subgroups with 4 numbers in sequence.

    If n>=2 then 2^n is divisible by four (2^(n-2) will be an integer). Since we remove n but add 0 it will always be possible t ...[text shortened]... I'm sure that someone can explain this more clearly than I did...
    The sums of the first powers would indeed be identical but if you look carefully you'll see that for n=3 (that is the numbers from 0 to 7), by using that method you'll end up with the groups A (0,3,4,7) and B (1,2,5,6) which indeed add up to the same number (14), but the sums of the squares is different 0 + 9 +16 +49 = 74 and 1 + 4 + 25 +36 = 66.
    And as I have stated in my example the groups are 1,2,4,7 and 0,3,5,6
    which again add to 14, and the sum of the squares of both groups are 70. (1+4+16+49 =70= 0 + 9 + 25 + 36).
  4. Standard member Palynka
    Upward Spiral
    18 Jan '05 17:18
    So that's why it sounded too easy to be true!
  5. 19 Jan '05 21:38 / 1 edit
    In order to help (confuse) anyone that is trying to find the rule for assigning the numbers in two groups, I will give the numbers in the groups for n=4 and n=5. So here they go:
    The groups are respectively:
    For n=4 A (0,3,5,6,9,10,12,15) and B(1,2,4,7,8,11,13,14). The sums of the first, second and third powers of both groups are equal...
    For n=5 A (0,3,5,6,9,10,12,15,17,18,20,23,24,27,29,30) and B(1,2,4,7,8,11,13,14,16,19,21,22,25,26,28,31). The sums are equal for all powers <= 4.

    One may easily see that all the numbers that were in the A and B groups for n=n1<n2 are again in the A and B groups for n=n2. However, the method by which the numbers are divided in these two groups is not that obvious, though it is quite simple in nature.