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Exclusively Minimum Groups (Posted on 2009-11-24) Difficulty: 3 of 5
Divide a set of 24 integers from 2 to 25 inclusively, with each of the integers occurring precisely once, into two mutually exclusive groups such that the absolute difference between the product of the integers in each group is the minimum.

No Solution Yet Submitted by K Sengupta    
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Factor that in... Comment 3 of 3 |

Charlie has apparently found a much lower solution than mine. I matched up several groupings which had the same products, so that either half of each group could be in either set, viz.:

(2 * 11) or 22...((3 * 7) or 21) ... (4*6) or 24 ...

(5*16) or (8*10);  (12 * 25) or (15 * 20)

Each side of each of these five obviously had the same prime factorization; I decided further testing could be more easily programmed.  This left only seven other integers to be tested:  9, 13, 14, 17, 18, 19, 23.  I programmed the comparison of all combinations of each of these between the two sets, c 2**7, less half to drop the symmetry. Using this list, I found the lowest difference in products was 19980 -- when 9, 13, 14, and 17 were in one set, and 18, 19, and 23 in the other.  I then multiplied this by the product of the members of each side of the five pairings I had excluded (1,555,286,261,760,000), since that would be the same for each set.  There must be a flaw in this somewhere, but I don't spot it.  (I split the task so that when I got to the actual computer part I could stay under 18 digits; then used the MS Calculator for the products.)   


  Posted by ed bottemiller on 2009-11-24 17:14:32
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