# Arrange the numbers 1 to 19 in the circles

     O---o---O
/ \     / \
o   o   o   o
/     \ /     \
O---o---O---o---O
\     / \     /
o   o   o   o
\ /     \ /
O---o---O


Arrange the numbers 1 to 19 in the circles, so all the rows of 3 numbers between O (big o) sums to 23

Example :

     a---b---c
/ \     / \
d   e   f   g
/     \ /     \
h---i---j---k---l
\     / \     /
m   n   o   p
\ /     \ /
q---r---s


a+b+c = a+e+j = c+f+j = 23, and so on....

Note :

• This puzzle is similar with this site, but with different sums
• There are 2 solutions, if we rule out reflections and rotations.

The sum from 1-19 equals 19*20/2 or 190.

If we examine the 12 sums, one for each line, it is observed that the 6 numbers on the vertices is used 3 times, and the number in the center is used 6 times.

So we have,

(sum of twelve numbers at the middle of the rows) + 3*(sum of six numbers at the vertices) + 6*(number at the center) = 23*12 = 276

Or,

2*(sum of six numbers at the vertices) + 5*(number at the center) = 86

The first term is always even, the number at the center must also be even. The sum of six numbers at the vertices is at least 1+2+3+4+5+6 = 21, so the number at the center can be at most 8. The possible numbers are 2,4,6 and 8.

Also, the number 19 has to be placed between 1 and 3, since it's the only way possible to make 23.

If the central number is 2, we know for sure that 1 isn't on a vertex, since that requires 20 between 1 and 2. Hence we have a contradiction.

If the central number is 4, the sum of the six numbers on the vertices must be 33. Since we need 1-x-3 (to place 19) on the outer edge, supposed 2 is not on a vertex, we must have a pair with the sum 21 to accompany 2. The maximum number we can have is 14 since 1+3+4+5+6+14=33, which implies that such pair cannot exist or the sum of the numbers on the vertices will exceed 33. So, 2 must be on a vertex. 2 cannot be adjacent to 1 (need 20 in between) or 3 (need 18 in between which is already taken in 1-18-4) so 2 must be either to the opposite of 1 or 3.

Suppose 5 is on a vertex, we have 33-(1+2+3+5) = 22 must be the sum of 2 non-adjacent numbers on the vertices (say x and y). 5 cannot be adjacent to 1 (5-17-1, 4-17-2). Neither x or y can be adjacent to 1 because x+y+1 = 23. Hence 5 is not on a vertex. 5 cannot be on the "radius" because then 14 must be on a vertex and it is impossible to fill in the blanks with distinct numbers. If 5 is on the outer edge, there exists a pair of number on adjacent vertices with the sum 18. Also 9 must also be on a vertex since 33-(1+2+3+18)=9, which is again, impossible to fill.

If the central number is 6, the sum of the six numbers on the vertices must be 28. Since we need 1-x-3 (to place 19), either 1-x-4 or 2-x-3 (to place 18) and either 1-x-5 or 2-x-4 (to place 17). So we either have either 2-17-4-18-1-19-3 on the outer edge,
which gives:

          1--19---3
/ \     / \
18  16  14  12
/     \ /     \
4--13---6---9---8
\     / \     /
17   15  7   5
\ /     \ /
2--11---10
or 1-19-3-18-2-17-4, which gives:

          1--19---3
/ \     / \
12  16  14  18
/     \ /     \
10--7---6---15--2
\     / \     /
5   9  13   17
\ /     \ /
8---11--4
The string 5-17-1-19-3-18-2 is also possible, however it is impossible to fill the rest of the numbers.

If the central number is 8, the sum of the six numbers on the vertices must be 23. It is only possible with 1,2,3,4,6,7. This is impossible since the number between 7(vertex) and 8(center) must be 23-7-8 = 8 which is a repeat.

Since all the possible cases are covered, it can be concluded that these are the only 2 distinct solutions.

• I have to post via phone so somebody please clear up the formatting mess for me. Thank you in advance. Oct 19, 2016 at 8:54