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Sum Rotation (Posted on 2004-05-11)

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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second part full solution

| Comment 15 of 17 |

(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) can´t 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 can´t 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.

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

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, THAT´S ENOUGH!! I DON´T KNOW WHERE I MADE A MISTAKE. TWO SOLUTIONS ONLY.

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

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