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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    
Rating: 3.2500 (4 votes)

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Solution second part full solution | Comment 15 of 16 |
(In reply to first part of full solution by pcbouhid)

Case IVa : G = 8, (B+C+E+F) = 17 and (A+D+H+I) = 20.

In (2), (H+I) = 8 or 18. If (H+I) = 18, (A+D) = 2, no way. If (H+I) = 8, (A+D) = 12. But in (1), (A+D) or (A+D+1) = 8. Eliminated.

Case IVb: G = 8, (B+C+E+F) = 26 and (A+D+H+I) = 11. 

In (2), (B+C) = A or (B+C+1) = A. If (B+C) = A, and A is at most 7, (E+F) >= 19, no way. And if (B+C+1) = A, (E+F) >=18, no way too. Eliminated.

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

In (2), (H+I) = 9 or 19. (H+I) canīt be 19, so (H+I) = 9, and (A+D) = 9.

Since (A+D) = 9, we have in (1), (B+E) = H (which means (C+F) = I) or (B+E+1) = H (which means (C+F) = 10+I).

(B+E) = H and (C+F) = I......(B+E+C+F) = (H+I). But (B+C+E+F) = 18 and (H+I) = 9. No way.

So, (B+E+1) = H and (C+F) = (10 + I) [or (C+F) = (19-H)].

Since A >= 3, we have (A,D) = (3,6), (4,5), (5,4), (6,3), (7,2) and (8,1).

A is greater than B and than C, and (B+C) = A or (B+C+1) = A.

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(A,D) = (3,6)..(B,C) = (1,2) or (2,1).

                       (B,C) = (1,2)... H = (E+2) = (17-F)...(E+F) = 15 so (E,F,H) = (8,7,10) or (7,8,9). Eliminated.   

                      (B,C) = (2,1)...H = (E+3) = (18-F)...(E+F) = 15 impossible because E must be less than 6 (see H).

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(A,D) = (4,5)...(B,C) = (1,3) or (3,1) or (1,2) or (2,1)

                       (B,C) = (1,3)...H = (E+2) = (16-F)...(E+F) = 14 so (E,F,H) = (8,6,10) or (6,8,8). Eliminated.

                       (B,C) = (3,1)...H = (E+4) = (18-F)....(E+F) = 14 so (E,F,H) = (8,6,12) or (6,8,10). Eliminated.

                       (B,C) = (1,2) and (2,1) eliminated as above.

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(A,D) = (5,4)..(B,C) = (2,3) or (3,2) or (1,3) or (3,1)

                       (B,C) = (2,3)...H = (E+3) = (16-F)...(E+F) = 13 so (E,F,H) = (7,6,10) or (6,7,9). Eliminated.

                       (B,C) = (3,2)...H = (E+4) = (17-F)...(E+F) = 13 so (E,F,H) = (7,6,11) or (6,7,10). Eliminated.

                       (B,C) = (1,3) and (3,1) eliminated .

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(A,D) = (6,3)...(B,C) = (5,1) or (1,5) or (4,2) or (2,4) or (4,1) or (1,4)

                       (B,C) = (5,1)...H = (E+6) = (18-F)...(E+F) = 12 impossible because E must be less than 4.

                       (B,C) = (1,5)...H = (E+2) = (14-F)...(E+F) = 12 so (E,F,H) = (8,4,10) or (7,5,9) or (5,7,7) or (4,8,6). Eliminated.

                       (B,C) = (4,2)...H = (E+5) = (17-F)...(E+F) = 12 impossible because E must be less than 4 (see H).

                       (B,C) = (2,4)...H = (E+3) = (15-F)...(E+F) = 12 so (E,F,H) =  (5,7,8)

                       (B,C) = (4,1)...H = (E+5) = (18-F)...(E+F) = 13 impossible because E must be less than 4 (see H).

                       (B,C) = (1,4)...H = (E+2) = (15-F)...(E+F) = 13 so (E,F,H) = (8,5,10) or (7,6,9) or (6,7,8) or (5,8,7) 

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(A,D) = (7,2)...(B,C) = (6,1) or (1,6) or (4,3) or (3,4) or (5,1) or (1,5)

                       (B,C) = (6,1)...H = (E+7) = (18-F)...(E+F) = 11 impossible because E must be 1 (to make H = 8).

                       (B,C) = (1,6)...H = (E+2) = (13-F)...(E+F) = 11 so (E,F,H) = (8,3,10) or (6,5,8) or (5,6,7) or (3,8,5). All but one eliminated.

                       (B,C) = (4,3)...H = (E+5) = (16-F)...(E+F) = 11, all possibilities eliminateds.

                       (B,C) = (3,4)...H = (E+4) = (15-F)...(E+F) = 11, all possibilities eliminated.

                       (B,C) = (5,1)...H = (E+6) = (18-F)...(E+F) = 12, no possibilities for (E,F,H).

                       (B,C) = (1,5)...H = (E+2) = (14-F)...(E+F) = 12, so (E,F,H) = (4,8,6).  

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(A,D) = (8,1)...(B,C) = (6,2) or (2,6) or (5,3) or (3,5) or (5,2) or (2,5) or (4,3) or (3,4) 

                       (B,C) = (6,2)...H = (E+7) impossible (E canīt be 1 or 2).

                       (B,C) = (2,6)...H = (E+3) = (13-F)...(E+F) = 10, all possibilities eliminated.

                       (B,C) = (5,3)...H = (E+6) = (16-F)...(E+F) = 10, no possibilities.

                       (B,C) = (3,5)...H = (E+4) = (14-F)...(E+F) = 10, no possibilities.

                       (B,C) = (5,2)...H = (E+6)....no values for E.

                       (B,C) = (2,5)...H = (E+3) = (14-F)...(E+F) = 11 no possibilities.

                       (B,C) = (4,3)...H = (E+5) = (16-F)...(E+F) = 11 no possibilities.

                       (B,C) = (3,4)...H = (E+4) = (15-F)...(E+F) = 11 no possibilities.

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========================================

A = 6, B = 2, C = 4, D = 3, E = 5, F = 7, G = 9, H = 8, I = 1.

A = 6, B = 1, C = 4, D = 3, E = 5, F = 8, G = 9, H = 7, I = 2.

A = 7, B = 1, C = 6, D = 2, E = 3, F = 8, G = 9, H = 5, I = 4.

A = 7, B = 1, C = 5, D = 2, E = 4, F = 8, G = 9, H = 6, I = 3.

========================================

624.......471       614......482    716....684     715....583

357.......258       358......157    238....135     248....146

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

981      639        972.....639     954....729     963.....729

     WRONG            RIGHT             WRONG         RIGHT

FOR A RAINNY DAY, THATīS ENOUGH!! I DONīT KNOW WHERE I MADE A MISTAKE. TWO SOLUTIONS ONLY.


  Posted by pcbouhid on 2005-08-12 20:18:04
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