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| 3x3 sums (Posted on 2004-05-15) |
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Fill a 3 by 3 grid with the digits 1 to 9 using the following five rules:
1. The sum of the top row is twice the sum of the center row.
2. The sum of the left column is twice the sum of the center column.
3. The sum of the right column plus twice the sum of the bottom row is equal to the sum of the whole grid.
4. The sum of the bottom row plus twice the sum of the right column is not equal to the sum of the whole grid.
5. The top row is the only row with both odd and even numbers.
Show that there is only one solution.
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Submitted by Axorion
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Rating: 3.0000 (3 votes)
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Solution:
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Let's call the columns left to right A, B and C. Likewise, the rows top to bottom D, E and F. The sum of the whole grid is 45 so the sums of A+B+C and D+E+F also equal 45. If rule 2: A=2B then 3B+C=45. If rule 3: 2F+C=45 then 3B=2F. This means B must be even.
If B=6 then A=12 and C=(45-A-B)=27. The highest sum we can get is 9+8+7=24 so B is not 6.
If B=10 then A=20, C=15, F=15, D=20 and E=10. This means that rule 4: 2D+F!=45 has been broken so B is not 10.
If B=12 then A=24, C=9, F=18, D=18 and E=9. Now if A=24 then the lowest number that can be at the left end of E is 7. If the other two digits in E are 1 and 2 then the lowest E can be is 10. E is not 9 therefore B is not 12.
Because A must be 24 or less and B is half of A, the only number left is B=8 so it is the only correct answer.
This gives us A=16, B=8 ,C=21, D=22, E=11 and F=12.
The only way B=8 and D=22 is for B=1+2+5 and D=5+8+9. This forces A=3+4+9 and C=6+7+8. Using rule 5: E=1+3+7 while F=2+4+6.
This leaves one possible solution.
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