October/Hallowe'en Special 2015 - Trick or Treat (Solution)

The answer is 80. Let the number of sweets carried by each be Gh, W, Go, V, S, and F respectively. We are given the following equations:

1: S = 2Go+2V
2: W = Gh+4V
3: S = 2F+28
4: F = (Gh+W+Go+V+S+F)/6
5: W+Go = Gh²
6: F = Go+3

From 4, 6F = Gh+W+Go+V+S+F so 5F = Gh+W+Go+V+S. Now, from 6, 5Go+15 = Gh+W+Go+V+S so 4Go+15 = Gh+W+V+S. Now, from 3 and 6, S = 2Go + 34. Since S = 2Go+2V, 2V = 34 so V = 17. Thus 4Go+15 = Gh+W+17+2Go+34 so 2Go = Gh+W+36. However, W = Gh+4V = Gh+68 so 2Go = Gh+(Gh+68)+36 = 2Gh+104. Thus Go = Gh+52.

We can now use 5. W+Go = (Gh+68)+(Gh+52) = 2Gh+120 so Gh² = 2Gh+120. We have the quadratic Gh²-2Gh-120 = 0. This factorises to (Gh-12)(Gh+10) = 0. Thus Gh = 12 or -10. However, Gh is obviously positive. Thus Gh = 12 and W = Gh+68 = 80.

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