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# cross sum

Started by brhan, Sep 29 2007 01:06 PM

2 replies to this topic

### #1

Posted 29 September 2007 - 01:06 PM

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.

### #2

Posted 29 September 2007 - 04:51 PM

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

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

### #3

Posted 01 October 2007 - 10:52 AM

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.

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.

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