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Assume AF is blank. Then AE is 7 and AG and AH are 8
and 9 in some order. Then DE is blank and the greatest sum available
for row D is 2+5+6 = 13. Therefore, AF is not blank. It cannot
be 1, because then there would need to be another 1 in column F, so AF is
2 and the remaining squares in column F are blank.
The digits in row B cannot be a 1 and 2, so it must be a single 3. Similarly, row C must contain a single 4. The digits in row A are either 2-5-8-9 or 2-6-7-9. It could not be the former, because then AE would have to be 5 and then there would need to be another 2 in column E. Therefore, the digits in row D are 1-5-8. The only way to make column E work is if AE is 6 and DE is 1. |
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AH is at least 7. If DH were 8, then AH would have to be 7, but there would need to be another 1 in column H. Therefore, DH is 5 and DG is 8. If AH were 9, then there would be to be another 2 in column H, so AH is 7 and AG is 9. We must then have BG=3 and CH=4 to complete the grid. |
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The solution is 6297 3 4 185. |
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All of the spots in row C and column E are filled, since
0+7+8+9 < 25. The sum of these seven numbers is at most 42
(3+4+...+9). If CE were 7 or less, then the other three numbers in
each of these two rows would be at least 18, which would yield a sum that
is too large. Therefore, CE is 8 or 9.
The 9 must be in row C, either at CE or CH. If it were at CH, then CE would be 8 and there would be a 1 in column H. The remaining 8 in row C could not be distributed 1-7 (the 1 is already used), 2-6 (we would need another 2 in column F), 3-5 (we would need 1's in both columns F and G), or any other way. So the 9 must be at CE. The sum of the remaining digits in row C and column A are 16 each, so the two digits in neither of these lines are 1 and 3. The only place to put a 2 would be at DE, so that the 3 is somewhere on row D. |
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8 must be at either AE or CH, and it cannot be at CH because then we would need another 2 in column H. BE is 6, so the 1 is somewhere on row B. The 7 must be at CH, which means that the 3 is at DH. The 5 must be at CG, which means that the 1 is at BG. Finally, the 4 is at CF. |
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The solution is 8 61 9457 23. |
Notes: I didn't look at this during the test at all. I was a little spooked at the difficulty of the example problem, especially since this was only worth ten points. Once I started looking at this after the test, I think that the second problem took me about fifteen minutes and the first problem another twenty.