Guest Posted September 29, 2007 Report Share Posted September 29, 2007 Put the numbers from 2 to 8 (without repeating) in each circle of the following diagram such that sum of the three numbers placed in a straight line (vertical, horizontal or diagonal) is 15. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 29, 2007 Report Share Posted September 29, 2007 Define: 1. x as the top of the trangle 2. y as the sum of a straight line Then, 2x + 35 = 3y (from the x down to three straight line) 35 - x = 2y (from the horizontal two straight line) => x = 5, y = 15 So, the remains is the sum of 10. that is, {2,8}, {3,7}, {4,6}. Fill in and make sure the straight line is 15. For example, 5 276 8 3 4 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted October 1, 2007 Report Share Posted October 1, 2007 Good work from Benson. The difficulty might be what number to place on the top circle. Let say X. Then, since the sum of the numbers from 2 to 8 is 35, ==> X + 2*15 = 35 (For the two horizontal row sums). ==> X = 5. Then as Benson said, pairing remaining numbers such that their sum is 10 -- 2 & 8; 3 & 7 and 4 & 6. Quote Link to comment Share on other sites More sharing options...
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Put the numbers from 2 to 8 (without repeating) in each circle of the following diagram such that sum of the three numbers placed in a straight line (vertical, horizontal or diagonal) is 15.
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