Question

I have already read What is an "external node" of a "magic" 3-gon ring? and I have solved problems up until 90 but this n-gon thing totally baffles me as I don’t understand the question at all.

So I take this ring and I understand that the external circles are 4, 5, 6 as they are outside the inner circle. Now he says there are eight solutions. And the eight solutions are without much explanation listed below. Let me take

9   4,2,3; 5,3,1; 6,1,2
9   4,3,2; 6,2,1; 5,1,3

So how do we arrive at the 2 solutions? I understand 4, 3, 2, is in straight line and 6,2,1 is in straight line and 5, 1, 3 are in a straight line and they are in clockwise so the second solution makes sense.

Questions

1. Why does the first solution 4,2,3; 5,3,1; 6,1,2 go anti clock wise? Should it not be 423 612 and then 531?
2. How do we arrive at 8 solutions. Is it just randomly picking three numbers? What exactly does it mean to solve a “N-gon”?

Answer 1

The first doesn’t go anti-clockwise. It’s what you get from the configuration

        4
         \
          2
         / \
        1---3---5
       /
      6

when you go clockwise, starting with the smallest number in the outer ring.

How do we arrive at 8 solutions. Is it just randomly picking three numbers? What exactly does it mean to solve a “N-gon”?

For an N-gon, you have an inner N-gon, and for each side of the N-gon one spike, like

        X
        |
X---X---X
    |   |
    X---X---X
    |
    X

so that the spike together with the side of the inner N-gon connects a group of three places. A “solution” of the N-gon is a configuration where you placed the numbers from 1 to 2*N so that each of the N groups sums to the same value.

The places at the end of the spikes appear in only one group each, the places on the vertices of the inner N-gon in two. So the sum of the sums of all groups is

 N
 ∑ k +  ∑{ numbers on vertices }
k=1

The sum of the numbers on the vertices of the inner N-gon is at least 1 + 2 + ... + N = N*(N+1)/2 and at most (N+1) + (N+2) + ... + 2*N = N² + N*(N+1)/2 = N*(3*N+1)/2.

Hence the sum of the sums of all groups is between

N*(2*N+1) + N*(N+1)/2 = N*(5*N+3)/2

and

N*(2*N+1) + N*(3*N+1)/2 = N*(7*N+3)/2

inclusive, and the sum per group must be between

(5*N+3)/2

and

(7*N+3)/2

• again inclusive.

For the triangle – N = 3 – the bounds are (5*3+3)/2 = 9 and (7*3+3)/2 = 12. For a square – N = 4 – the bounds are (5*4+3)/2 = 11.5 and (7*4+3)/2 = 15.5 – since the sum must be an integer, the possible sums are 12, 13, 14, 15.

Going back to the triangle, if the sum of each group is 9, the sum of the sums is 27, and the sum of the numbers on the vertices must be 27 - (1+2+3+4+5+6) = 27 - 21 = 6 = 1+2+3, so the numbers on the vertices are 1, 2 and 3.

For the sum to be 9, the value at the end of the spike for the side connecting 1 and 2 must be 6, for the side connecting 1 and 3, the spike value must be 5, and 4 for the side connecting 2 and 3.

If you start with the smallest value on the spikes – 4 – you know you have to place 2 and 3 on the vertices of the side that spike protrudes from. There are two ways to arrange the two numbers there, leading to the two solutions for sum 9.

If the sum of each group is 10, the sum of the sums is 30, and the sum of the numbers on the vertices must be 9. To represent 9 as the sum of three distinct numbers from 1 to 6, you have the possibilities

1 + 2 + 6
1 + 3 + 5
2 + 3 + 4

For the first group, you have one side connecting 1 and 2, so you’d need a 7 on the end of the spike to make 10 – no solution.

For the third group, the minimal sum of two of the numbers is 5, but 5+6 = 11 > 10, so there’s no place for the 6 – no solution.

For the second group, the sums of the sides are

1 + 3 = 4  -- 6 on the spike
1 + 5 = 6  -- 4 on the spike
3 + 5 = 8  -- 2 on the spike

and you have two ways to arrange 3 and 5, so that the group is either 2-3-5 or 2-5-3, the rest follows again.

The solutions for the sums 11 and 12 can be obtained similarly, or by replacing k with 7-k in the solutions for the sums 9 resp. 10.

To solve the problem, you must now find out

• what it means to obtain a 16-digit string or a 17-digit string
• which sum for the groups gives rise to the largest value when the numbers are concatenated in the prescribed way.

(And use pencil and paper for the fastest solution.)