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Rotate the Outer (Posted on 2009-08-30) Difficulty: 2 of 5

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analytical solution | Comment 1 of 2

Let the grid be
abc
dSe
fgh
with a,b,c,d,e,f,g,h being consecutive integers not neccisarily in order.
now we have
S=2(a+c+f+h)+b+d+e+g (1)
but if we rotate then we have
bce
aSh
dfg
and then we have
S=2(b+e+g+d)+(c+a+h+f) (2)
taking (1)-(2) we end up with
a+c+f+h=b+d+e+g (3)
now let the consecutive integers be
x,x+1,x+2,x+3,x+4,x+5,x+6,x+7
by taking
a=x c=x+7 f=x+1 h=x+6 and
b=x+2 d=x+5 e=x+3 g=x+4 then we have both sides of (3)
equal to 4x+14

In general, all you need for this to work is to have the sum of the corner numbers be equal to the sum of the orthogonal numbers.


  Posted by Daniel on 2009-08-30 14:05:42
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