Eight Circles, Sum is 13

Date: 09/25/2002 at 11:21:37
From: Trevor Callahan
Subject: Sum of Thirteen

Place the numbers 1,2,3,4,5,6,7,8 in the eight circles so that the sum 
of the numbers in any line equals 13.

          O     O     O

          O           O  

          O     O     O

There are two different solutions.

          O     O     O

          O           O

          O     O     O

I am getting nowhere. Please help soon.

Date: 09/26/2002 at 01:35:11
From: Doctor Greenie
Subject: Re: Sum of Thirteen

Hi, Trevor -

Thanks for sending this nice little puzzle to us here at Dr. Math.  
I had some fun and got some good mental exercise in reasoning this one 
out.

The given information says the sum of the numbers in each row of 
three must be 13. The total of all four of these rows is then 13*4=52.

The sum of the numbers we are to use: 1, 2, ..., 7, and 8, is 36.

In finding the sum of the four rows of three, the numbers in the 
corners of the array of circles get counted twice. This means that the 
difference between the sum of the numbers in all four rows of three 
and the sum of the eight numbers we are using is the sum of the four 
numbers in the corners of the array.

So we know the following:

  the sum of the four corner numbers is  52 - 36 = 16

Now, with the sum of the three numbers in each row being the odd 
number 13, each of these four rows must contain either (a) three odd 
numbers or (b) two even numbers and one odd number.  Since the whole 
array must contain four even and four odd numbers, the only possible 
arrangements of evens and odds in the array are the following two 
(where "E" and "O" denote even or odd numbers, respectively):

    E   O   E        O   O   O

    O       O   or   E       E

    E   O   E        E   O   E

However, we can rule out the first possibility, because in that 
configuration, the sum of the four corner numbers is 2+4+6+8=20; and 
we have determined that the sum of the four corner numbers must be 16.

So the configuration of even and odd numbers in the array is the 
second pattern shown above.

Now, the sum of the numbers in the two vertical rows must be 13*2=26; 
and the sum of the four even numbers is 20; therefore, the sum of the 
two odd numbers in the two vertical rows must be 6, so those odd 
numbers are 1 and 5.  And then the other odd number in the top row 
must be 7 to make the sum in that row 13; so the odd number in the 
bottom row is 3.

So now we have determined that the configuration is

    1   7   5

    E       E

    E   3   E

From here, you can determine the possible placements of the even 
numbers in their four slots by working with the required sum of 13 in 
either vertical row or in the bottom row; you will see that there are 
two solutions in the pattern we have established:

    1   7   5          1   7   5

    4       6    or    8       2

    8   3   2          4   3   6

These are the two solutions to your problem.

I hope this helps.  Please write back if you have any further 
questions about any of this.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/