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Equal sums on a circle (Posted on 2013-11-26) Difficulty: 2 of 5
Place the numbers 1 to 10 around a circle such that the sum of any two adjacent numbers is equal to the sum of the pair of numbers directly opposite. In how many ways can this be done? Can this be done with 1 to 8? With 1 to 6?

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution computer solution Comment 1 of 1

DECLARE SUB place ()
DIM SHARED n, nbr
n = 10
DIM SHARED used(n), cir(n), ct

CLS
place

PRINT ct

SUB place
  nbr = nbr + 1
  IF nbr <= n / 2 + 1 THEN
     IF nbr = 1 THEN hi = 1:  ELSE hi = n
     FOR i = 1 TO hi
       IF used(i) = 0 THEN
          cir(nbr) = i
          used(i) = 1
          place
          used(i) = 0
       END IF
     NEXT
  ELSE
     t = cir(nbr - n / 2) + cir(nbr - 1 - n / 2)
     newnum = t - cir(nbr - 1)
     IF newnum > 0 AND newnum <= n THEN
       IF used(newnum) = 0 THEN
         cir(nbr) = newnum
         used(newnum) = 1
          IF nbr = n THEN
           t = cir(1) + cir(n)
           t2 = cir(1 + n / 2) + cir(n - n / 2)
           IF t = t2 AND cir(2) < cir(n) THEN
             FOR i = 1 TO n
               PRINT cir(i);
             NEXT
             PRINT
             ct = ct + 1
           END IF
          ELSE
           place
          END IF
         used(newnum) = 0
       END IF
     END IF
  END IF
  nbr = nbr - 1
END SUB

finds the following set of 24 solutions:

1  4  5  8  9  2  3  6  7  10
1  4  5  10  7  2  3  6  9  8
1  4  7  6  9  2  3  8  5  10
1  4  7  10  5  2  3  8  9  6
1  4  9  6  7  2  3  10  5  8
1  4  9  8  5  2  3  10  7  6
1  6  3  8  9  2  5  4  7  10
1  6  3  10  7  2  5  4  9  8
1  6  7  4  9  2  5  8  3  10
1  6  9  4  7  2  5  10  3  8
1  7  3  9  5  6  2  8  4  10
1  7  3  10  4  6  2  8  5  9
1  7  4  8  5  6  2  9  3  10
1  7  4  10  3  6  2  9  5  8
1  7  5  8  4  6  2  10  3  9
1  7  5  9  3  6  2  10  4  8
1  8  2  9  5  6  3  7  4  10
1  8  2  10  4  6  3  7  5  9
1  8  3  6  9  2  7  4  5  10
1  8  4  7  5  6  3  9  2  10
1  8  5  4  9  2  7  6  3  10
1  8  5  7  4  6  3  10  2  9
1  9  2  8  5  6  4  7  3  10
1  9  3  7  5  6  4  8  2  10

All are presented starting at the 1. So as to eliminate reflections, only those with a lower number to the 1's right than to its left (at the end of each list) are shown; there would be 48 solutions if the reverse sequences were counted.

By changing the value of n and rerunning we find two solutions (4 if you count reflections) for n=6:

1  4  5  2  3  6
1  5  3  4  2  6

but none for n = 8.

There are also no solutions for n = 12.

Going to n=14, there are 720 solutions (1440 counting reversals).


  Posted by Charlie on 2013-11-26 12:13:40
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