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One triplet out (Posted on 2015-06-16) Difficulty: 3 of 5
Given set of fifteen integers (1 to 15) .

Erase 3 numbers so that remaining integers can be arranged as a 3 by 4 array in which the sum of the numbers in each row is a certain Sr and in each of the columns a certain Sc.

Present the triplet you have chosen and one of the possible arrangements.

D4. Bonus question:
How many distinct solutions are there?

No Solution Yet Submitted by Ady TZIDON    
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Here's one | Comment 1 of 3
I'm guessing there are a great many solutions.  Here's one I found pretty quickly:
14  1  8  5 
 3 13  2 10
 4  7 11  6
The columns sum to 21 and the rows sum to 28.

All I did was note that the remaining numbers must sum to a multiple of 12 so as to be divisible by both 3 and 4.  

Since the numbers 1 to 15 sum to a multiple of 12, the numbers removed must also sum to a multiple of 12.

I chose to remove 15, 12, 9 because they have a large sum are are spaced well apart (I'm not sure why, really)

The 4 columns then had to sum to 21.  There are only a dozen partitions that are possible.  This is the only quadruplet that doesn't repeat numbers.  Then I arranged them to make the row sums all 28.

  Posted by Jer on 2015-06-16 13:27:52
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