我也跟不上。只能再敲一边。

let us use a,b,c,d,e,f,g,h and i be numbers as the following configuration:

a----b----c

d----e----f

g----h----i

we know that S=a+c+g+h=d+b+h+f=a+b+d+e=b+c+e+f=d+e+g+h=e+f+h+i.

45=e+ (a+c+g+i) + (b+d+h+f) =>45=e+2S;   ----(1)

2S=(a+d+b+e) + (e+h+i+f) = 2e + (a+i) + (d+b+h+f)=(a+i) + 2e + S =>a+i+2e=S;

similarly c+g + 2e=S; a+i + 2e = S; => (c+g+a+i) +4e= 2S => 4e=S; ---(2)

from (1) and (2), we have e=5 and S =20.

Now 5+9=14, in the remaining numbers only 2+4=6, so 9 must be on the corner and surrounded by 2 and 4. Due to the symmetry, we can have:

a----b----c

d----5----4

g----2----9

in the remaining numbers, only 3+8 =11, and only 6+7=13, so 5, 4, 3, and 8 are in the same small square. And 6, 7, 2, and 5 are in another small square. It is easy to see we must have a=1 across the diagnal from 9. Only 6+8=14, so 6, 8, 1,and 5 are in the same small square.

The only solution would be the following:

1----8----3

6----5----4

7----2----9

所有跟帖： 

• 哇,高.不过对小孩,说理不如背诀.理在其中.特编几句. -jinjing- ♀ (242 bytes) () 11/20/2012 postreply 05:49:32

• 哇,高.可以简单证明对角对边的和是10 -童关仙子- ♂ (0 bytes) () 12/20/2012 postreply 08:30:17