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A quite even partition (Posted on 2005-08-28) Difficulty: 3 of 5
Can you partition the numbers 1, 2, 3, ... nē in n separate subsets, each with n numbers, all subsets having the same sum?

See The Solution Submitted by Federico Kereki    
Rating: 4.3333 (3 votes)

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Solution | Comment 2 of 18 |
 
Here is the procedure with N = 4.
Fill a 4x4 array, say A, with the numbers 1-16:
  A | 0| 1| 2| 3|
  --+--+--+--+--+
  0 | 1| 2| 3| 4|
  --+--+--+--+--+
  1 | 5| 6| 7| 8|
    +--+--+--+--+
  2 | 9|10|11|12|
    +--+--+--+--+
  3 |13|14|15|16|
    +--+--+--+--+
Cycle row R left R places
  A | 0| 1| 2| 3|
  --+--+--+--+--+
  0 | 1| 2| 3| 4|
  --+--+--+--+--+
  1 | 6| 7| 8| 5|
    +--+--+--+--+
  2 |11|12| 9|10|
    +--+--+--+--+
  3 |16|13|14|15|
    +--+--+--+--+
     34 34 34 34
The columns add up to 34 = [N^2*(N^2+1)/2]/N.

An element of the array A is
  A(R,C) = N*R + mod(R+C,N) + 1
Summing a column,
  sum(A(R,C),R=0 to N-1)
    = sum(N*R + mod(R+C,N) + 1,R=0 to N-1)
    = sum(N*R,R=0 to N-1) +
      sum(mod(R+C,N),R=0 to N-1) +
      sum(1,R=0 to N-1)
    = N*sum(R,R=0 to N-1) +
      sum(R,R=0 to N-1) +
      N
    = (N+1)*sum(R,R=0 to N-1) + N
    = (N+1)*[N*(N-1)/2] + N
    = [N^2*(N^2+1)/2]/N.
 

  Posted by Bractals on 2005-08-28 16:39:14
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