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Sum Rotation (Posted on 2004-05-11) Difficulty: 3 of 5
The numbers 1 through 9 are arranged in a 3 x 3 grid so that each number is in the grid exactly once. They are arranged such that the top row plus the middle row gives the bottom row. If the grid forms another such addition when it is rotated 90 degrees to the left, what is its composition?

(Note: The numbers don't flip, for example 6 doesn't turn into 9.)

No Solution Yet Submitted by Gamer    
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first part of full solution | Comment 14 of 16 |

1) ABC + DEF = GHI

2) CFI + BEH = ADG

From (1), G is greater than A and than D, and from (2), A is greater than C and than B (thus A is at least 3). So G must be at least 5.

From (1), (A + B + C + D + E + F) congr to (G + H + I) mod 9.

From (2), (C + F + I + B + E + H) congr to (A + D + G) mod 9.  

Summing up:

(A+2B+2C+D+2E+2F+H+I) congr (A+D+2G+H+I) mod 9.

(2B + 2C + 2E + 2F) congr (2G) mod 9.

2(B + C + E + F) congr (2G) mod 9.

I) G = 5....(B+C+E+F) = 14 or 23....(A+D+H+I) = 26 or 17.

II) G = 6...(B+C+E+F) = 15 or 24...(A+D+H+I) = 24 or 15.

III) G = 7..(B+C+E+F) = 16 or 25..(A+D+H+I) = 22 or 13.

IV) G = 8...(B+C+E+F) = 17 or 26..(A+D+H+I) = 20 or 11. 

V) G = 9....(B+C+E+F) = 18...........(A+D+H+I) = 18.

Remembering that A, B, C and D are less than G (G>=5):

-------------------------------------------------------------

Case Ia: G = 5, (B+C+E+F) = 14 and (A+D+H+I) = 26.

(A+D) is at most 7, so (H+I) >= 19. Eliminated. 

Case Ib: G = 5, (B+C+E+F) = 23 and (A+D+H+I) = 17.

In (2), (H+I) = G or 10+G, that is 5 or 15. If (H+I) = 5, (A+D) = 12, no way, and if (H+I) = 15, (A+D) = 2, no way too. Eliminated.

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Case IIa: G = 6, (B+C+E+F) = 15 and (A+D+H+I) = 24.

In (2), (H+I) = G or 10+G, that is 6 or 16. If (H+I) = 6, (A+D) = 18, no way, and if (H+I) = 16, (A+D) = 8. But in (1), (A+D) or (A+D+1) = G = 6. Eliminated.

Case IIb: G = 6, (B+C+E+F) = 24 and (A+D+H+I) = 15.

In (2), (B+C) = A or (B+C+1) = A. If (B+C) = A, and A is less than G(6), so (B+C) is at most 5, so (E+F) >=19, no way. If (B+C+1) = A, (E+F)>=18, no way too. Eliminated.

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Case IIIa: G = 7, (B+C+E+F) = 16 and (A+D+H+I) = 22.

In (2), (H+I) = 7 or 17. If (H+I) = 7, (A+D) = 15, which is impossible since A and D are less than 7(G). If (H+I) = 17, then (A+D) = 5. But in (1), (A+D) or (A+D+1) = G = 7. Eliminated.

Case IIIb: G = 7, (B+C+E+F) = 25 and (A+D+H+I) = 13.

In (2), (B+C) = A or (B+C+1) = A. If (B+C) = A, and A is at most 6 (A<G), then (E+F) >=19, no way. If (B+C+1) = A, (E+F) >= 20, no way too. Eliminated.

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The other 2 hypotheses in next posting.

 


  Posted by pcbouhid on 2005-08-12 17:10:39
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