Arrange the numbers from 1 to 19 in the circles shown in the picture above in such a way that, every one of twelve lines containing three circles add up to exactly 23. (Ignore the line segments with the hub of the diagram in the middle of the line.)

  • 2
    $\begingroup$ Do you want the total to be 23 or not? Also, there are actually 15 lines containing 3 circles. $\endgroup$
    – Amit Naidu
    Feb 17, 2018 at 7:29

2 Answers 2


The possible answers are:


and put 7-12 on one of the 6 middle circles.

How I achieve that is:

One of the outer line must be 1-19-3 and other line must be one of the following: 4-18-1 and 3-18-2. The rests outer line is just a simple trial-and-error.
Then I notice that every corners are summed up to 7 with its opposite, so putting 16 in the center will be a complete solution.

  • $\begingroup$ If I understood this question right, I'm not sure how this would work... so for example on the left horizontal spoke you need $16+5+n=23$, then $n$ must be $2$, but you already used $2$... Same for the next spoke in the counterclockwise direction; needs a $3$ in the blank spot but you've already used it. Am I missing something? $\endgroup$
    – Guest
    Feb 17, 2018 at 17:36
  • $\begingroup$ I agree with Guest. The question says there are twelve lines, not nine, suggesting that each spoke is being counted as a separate line. $\endgroup$
    – hexomino
    Feb 17, 2018 at 18:31
  • $\begingroup$ whooops, I agree with you. I had a big mistake here, I'm gonna update my answer then. $\endgroup$
    – athin
    Feb 18, 2018 at 1:10

I am pretty sure there are only 2 solutions, ignoring the symmetric mirror cases.

02 11 10 17 15 07 05 04 13 06 09 08 18 16 14 12 01 19 03


08 11 04 05 09 13 17 10 07 06 15 02 12 16 14 18 01 19 03


It is hard to explain and could be confusing to follow unless you have worked through the problem a bit. Starting with the large numbers, 19 has only one case: 1-19-3 and can only go on the perimeter.
18 can only be in 3 positions, in the row adjacent to the 19. Also, the two middle numbers in that row must differ by 2, because of the 1 and 3.
Similarly 17 can only take about 7 slots. Then the center number can narrowed down to 4-8 because it must combine with at least 6 completely different pairs of numbers to add up to 23. Another insight that is useful during trial and error is that the 6 perimeter slots which are not connected to the center spokes follow a different rule. Only specific numbers can go in those.
The 6 corner vertices cannot have numbers >14 because they all require 3 different pairs of numbers to be available to share the vertex number for a common sum.


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