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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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a chosen triplet Comment 3 of 3 |
Let the sum of each row, given that each row sums to the same total, be represented as R. Let the sum of each column, given that each column sums to the same total, be represented as C.
3R = 4C. One can deduce that the total of the array is divisible by 12 (3 and 4).

The smallest sum of the fifteen less three integers is 78
(= 1+2+3+4+5+6+7+8+9+10+11+12), and
the largest sum of the fifteen less three integers is 114
(= 4+5+6+7+8+9+10+11+12+13+14+15).
The integers within this range that are divisible by 12 are
84, 96, and 108.

Let A be the sum of the thirteen integers.
If A =  84, then R = 28 and C = 21.
If A =  96, then R = 32 and C = 24.
If A = 108, then R = 36 and C = 27.

The three integers (triplet) of the fifteen integers (1 to 15) eliminated must be divisible by 12, as the sum of the fifteen integers totals 120, which is divisible by 12, and as A is deduced to be divisible by 12.

These triplet possibilities for A = 108 are:
{1,2,9}, {1,3,8}, {1,4,7}, {1,5,6}, {2,3,7}, {2,4,6}, and {3,4,5}.

A solution exists for triplet {1,4,7}, other solutions may exist:

15 11  8  2 =  36
 3  6 14 13 =  36
 9 10  5 12 =  36
-----------   ---
27 27 27 27 = 108

  Posted by Dej Mar on 2015-06-16 18:02:23
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