This is a fun puzzle from John Bassey at Puzzle Sphere.
“The diagram shows a heptagon with three circles on each side. Some circles already have the numbers 8 to 14 filled in, while the remaining circles need to be filled with the numbers 1 to 7. Each circle must contain one number, and the sum of the numbers in every set of three circles along a line must be the same. Arrange the numbers!!!”
See Fill in the Blanks for a solution.
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Let the heptagon be ABCDEFG.
Let the numbers at the vertices be A,B,etc.
The sum of 7 side sums will be (8+10+..+14)+(1+2+..+7)+(1+2+..+7)
= (1+2+3..+14) + (1+2..+7)
= 14*15/2 + 7*8/2
= 105 + 28 = 133
Hence each side sum is 133/7 = 19.
The side AB has 14 as midpoint number. It can have numbers either 1,4 or 2,3 at A and B.
The side GA has 8 as midpoint number. It can have numbers either 4,7 or 5,6 at G and A.
From above, A can be 4.
Rest is easy.
B = 19 – 14 -4 = 1
C = 19 – 13 – 1 = 5
D = 19 – 12 – 5 = 2
E = 19 – 11 – 2 = 6
F = 19 – 10 – 6 = 3
G = 19 – 9 – 3 = 7
The 7 numbers at the 7 vertices A,B,C,D,E,F,G are 4,1,5,2,6,3,7 respectively.
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