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| B and C must be either 2 and 3 or 2 and 4 so that D is a single-digit number different from B and C. The only way to work with A is A=1, B=2, C=3, D=6. The only way to make G+H = J with the remaining digits is J=9 and G and H to be 4 and 5 in some order. Therefore, E and F are 7 and 8 in some order. The only way to make D+E = F+G work is E=7, F=8, G=5, H=4. |
| The solution is 1 2 3 6 7 8 5 4 9. |
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| E is at least 6 (since E = F+G+H) but at most 8 (since E+G
= J).
If E were 6, then F, G, and H would have to be 1, 2, and 3 in some order. To make A+B=C work with the remaining numbers, C would be 9, A and B would be 4 and 5 in some order, and D and J would be 7 and 8 in some order by elimination. So B+C+D would be between 20 and 22, which would not satisfy 6F. If E were 7, F, H, and H would be 1, 2, and 4 in some order. From the remaining numbers, B+C+D would be between 14 and 23, so F=2 and B, C, and D would be 3, 5, and 6 in some order. But then there is no way to satisfy A+B=C. |
| So E is 8. This makes G = 1 and J = 9. F could not be 3 or more, because B+C+D could not be large enough, so F=2 and H=5. A=4 because the other three digits sum to 16, so B=3, C=7, and D=6. |
| The solution is 4 3 7 6 8 2 1 5 9. |
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| G = B+C+D = 2B+A+D. B is at most 3.
If B were 3, then A and D would be 1 and 2 in some order and G would be 9. But there would be no way to satisfy 3E+F = 9. If B were 2, then A, D, and E would all be at most 4. Therefore, F and H would each be at least 5, which would make J more than 9. |
| So B=1. G is at least 7, and G = E+F = 2+A+D.
If G were 7, then A and D would be 2 and 3 in some order, but there would be no way to satisfy G = E+F. If G were 8, then A and D would be 2 and 4 in some order. E and F could be 3 and 5, but there would be no way to satisfy A+1 = C. |
| So G=9. The 8 can only go at J. The 7 can only go at E, so F=2, and H=6. C and D must be 5 and 3, so A=4, C=5, and D=3. |
| The solution is 4 1 5 3 7 2 9 6 8. |
Notes: I didn't look at any of these during the test. It seems to me that only the first one was easy enough to be worth ten points, as the other two took me approximately twenty minutes apiece afterwards.