## Magic Octagon

 ``` p a b o c n d m q e l f k g j i h``` Arrange the numbers 1-17 at positions a-q in the octagon, such that the eight sides (i.e. pab, bcd, ...) and the eight lines through the center (i.e. aqi, bqj, ...) all have the same sum. For ease in comparing solutions, let 'a' be the smallest of (a,c,e,g,i,k,m,o) and 'b' be less than 'p'.
Source: Denis Borris.
Solutions were received from Dennis Molter, Denis Borris, Radu Ionescu, Claudio Baiocchi, Joseph DeVincentis, and Philippe Fondanaiche.  The common sum is 27 and the center number must be 9.  There are four possible solutions:
 ``` 16 1 10 6 14 5 3 7 9 11 15 13 4 12 8 17 2``` ``` 14 1 12 2 10 11 5 3 9 15 13 7 8 16 6 17 4``` ``` 15 2 10 11 13 1 4 12 9 6 14 17 5 7 8 16 3``` ``` 17 3 7 8 14 2 6 13 9 5 12 16 4 10 11 15 1```
There are several ways to find the values of q and the common sum S.  Here's one: Adding four non-adjacent sides, and two rows through the center will use all entries once, and q twice.  So 6S-q = 17*18/2 = 153.  q must be 3, 9, or 15.  If we instead add all eight lines through the center, we use all entries once and q eight times, so 8S-7q = 153.  q must be 1, 9 or 17.  So q=9, and S=27.
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