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Sum of cubes and squares (Posted on 2011-08-28) Difficulty: 3 of 5
Consider N positive integers which are not necessarily distinct such that their sum of cubes is equal to 2011 (base ten) and their sum of squares is a perfect square.

Determine the smallest value of N for which this is possible. What is the next smallest value of N?

No Solution Yet Submitted by K Sengupta    
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solution | Comment 1 of 2

let xk be the number of occurences of the integer k in the set
then we need sum xk*k^3 = 2011 and sum xk*k^2 = s^2 for some integer s.  12<=2011^(1/3)<13 so the largest element possible is 12.  This gives for a simple exhaustive search that gives the following three sets of minimal length 10
(x1,x2,x3,x4,x5,x6,x7,x8,x9,x10,x11,x12)
(0,0,3,4,0,1,0,0,2,0,0,0)
(0,0,4,2,0,2,1,0,0,1,0,0)
(2,1,3,3,0,0,0,0,0,0,0,1)
and the next longest has length 11 with 4 sets
(2,0,3,1,0,4,0,0,0,1,0,0)
(2,1,0,3,2,1,1,0,0,1,0,0)
(0,3,4,0,0,0,1,3,0,0,0,0)
(0,4,2,0,1,1,1,1,1,0,0,0)

In total, there are 53340 sets


  Posted by Daniel on 2011-08-28 19:06:43
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