9514 1404 393
Answer:
[[. . 3 6 .][1 3 4 9 .][9 2 . 7 1][. 1 2 8 3][. 4 9 . . ]]
Step-by-step explanation:
It is helpful to remember that the first four digits {1, 2, 3, 4} total 10, and the last four digits {6, 7, 8, 9} total 30. It is also worthwhile to note that the smallest sum of two digits is 3 = 1+2, and the largest sum of 2 digits is 17 = 8+9. Some sums can only be made one way: 4 = 1+3, 16 = 7+9.
Here, we will refer to rows 1-5 from the top down, and columns 1-5 from the left to right.
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Based on the above, column 5 can only be made one way: 1+3. Column 4 can only be made one way: 9+8+7+6. Then the choices for row 3B will be 1+7 or 5+3. We know it can't be 5+3 because 5 is not part of column 4. This makes our first fill-in (row, column) be (3,4)=7, (3,5)=1, (4,5)=3.
The sum of the remaining cells in row 4 must be 14-3 = 11, and cell (4,4) must be 6, 8, or 9. It cannot be 9, because that would require the remaining two cells to have a sum of 2, which they cannot. So, cell (4,4) is 6 or 8.
Cell (1,4) cannot be 9, either, because we cannot use 0 in cell (1,3). Then the only place for 9 in column 4 is in cell (2,4)=9.
With cell (1,4) being restricted to 6 or 8 (the two remaining numbers in column 4, cell (1,3) must be 3 or 1, and cell (2,3) must be 4 or 6. It cannot be 6 because that would require a sum of 2 in the first two cells of row 2, so (2,3)=4, (1,3)=3, and (1,4)=6. This makes (4,4)=8 and the remaining two cells of row 4 must be 1 and 2. 1 cannot go in (4,3), so we must have (4,3)=2 and (5,3)=9, (5,2)=4, (4,2)=1. That leaves 2 and 3 for rows 2 and 3 of column 2. The first two cells of row 2 must total 4, so must be 1 and 3. Cell (2,2) cannot be 1, so must be 3, leaving (2,1)=1, (3,1)=9, (3,2)=2 and that completes it.
The fill-ins are ...
[. . 3 6 .]
[1 3 4 9 .]
[9 2 . 7 1]
[. 1 2 8 3]
[. 4 9 . . ]
