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 A quite even partition (Posted on 2005-08-28)
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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