1. ) It is given that A1 is 1 and A2 is 2.
2. ) Since numbers increase in value across each row and down each column, the largest number must be in the bottom-right corner. Therefore, 25 is in cell E5.
3. ) 3 must either be placed in cell A3 or B1 to keep numbers increasing in value across each row and down each column. But according to cell E5, every column contains exactly one pair of numbers whose values are consecutive. Since 1 and 2 already appear in Column A, 3 cannot be placed in A3. Therefore, 3 must be in cell B1.
4. ) Since Row 5 cannot have any consecutive numbers in adjacent cells according to the clue in cell B1, 24 cannot be placed in D5. The only other cell in which it can be placed is E4 to keep numbers increasing in order across each row and down each column.
5. ) 23 must either be placed in cell D5 or E3 to keep numbers increasing in value across each row and down each column. There is already a pair of consecutive numbers in Column E, so 23 cannot be placed adjacent to 24 in cell E3. Therefore, 23 must be placed in cell D5.
6. ) It is given by the clue in cell E4 that 10 is in cell D1.
7. ) To keep numbers increasing across each row and down each column, 22 must be placed in either cell C5, D4, or E3. Since there are no consecutive numbers in Row 5, it cannot be placed in cell C5 next to 23. It also cannot be placed in cell E3 since it is an even number, and cell D5 states that Row 3 only contains odd numbers. By default, 22 must go in cell D4.
8. ) According to cell D4, Row 1 sums to 31. That means the remaining cells (C1 and E1) must sum to 17. Furthermore, cell B1 states that there is a pair of consecutive numbers adjacent in the row. The only ways to satisfy both clues while keeping numbers increasing in value across the row are to place either (4 and 13) or (6 and 11) in cells C1 and E1, respectively.
9. ) Cell A2 states that Row E sums to 100. Since E1 must be either 11 or 13 from step 8, the series of cells E1, E2, and E3 must be either (11, 19, and 21), (13, 18, and 20), or (13, 17, and 21), respectively. Since Row 3 contains only odd numbers, the middle option is eliminated, and the only number that can be placed in cell E3 is 21. E1 and E2 are either (11 and 19) or (13 and 17), respectively.
10. ) According to cell D1, Column C does not contain any prime numbers. According to cell D5, Row 3 contains only odd numbers. Therefore, cell C3 must be an odd composite number. The only odd composite numbers not yet placed in the grid are 9 and 15.
11. ) Row 3 sums to 65 according to cell E3. If 9 were placed in cell C3, then the only small odd numbers remaining to place in A3 and B3 are 5 and 7, respectively. To make the row sum to 65, D3 would have to be 23, which was already placed in the grid and is larger than 21 in E3. Therefore, 9 cannot be placed in cell C3, which means that C3 must contain 15.
12. ) The remaining cells in Row 3 must be odd numbers not yet placed in the grid and must increase in order from left to right. That means that 5, 7, 9, 11, and 13 could be placed in cells A3 and B3, and either 17 or 19 must be placed in cell D3. The whole row must sum to 65 and its current sum is 36, so the three remaining cells must sum to 29. The only way to make this sum correctly add up is to use the lowest possible values: 5, 7, and 17. Therefore, 5 is in cell A3, 7 is in B3, and 17 is in D3.
13. ) Now that 17 has been placed in the grid, it cannot be used again, which means that of the two options from Step 9 to complete Column E, the only possible solution is for E1 to be 11 and E2 to be 19.
14. ) To make Row 1 sum to 31, C1 must be 6.
15. ) In order to keep numbers increasing across each row and down each column, 4 must be placed in cell B2 and 20 in cell C5.
16. ) Column C is the only column to not yet contain a pair of consecutive numbers. The only numbers that have not yet been placed that could fit this criterion are 14 in cell C2 or 16 in C4. If 14 were in C2, then all of the remaining numbers smaller than 14 (8, 9, 12, and 13) would have to be placed in cells A4, A5, B4, and B5. However, no consecutive numbers can be placed in Rows 4 or 5 (clue from B1), nor can a second pair be placed in Columns A or B (clue from E5). There are no possible arrangements for the four numbers in those four cells, so the scenario cannot exist. Since 14 cannot be placed in C2, then 16 must be placed in C4.
17. 18 must be placed in B5 in order to keep the numbers increasing in value across each row and down each column.
18. ) Cell C4 says that there is a second perfect square in Row 4. The only perfect square not yet placed is 9, so it must be in either A4 or B4. If 9 were placed in B4, then 8 would have to be placed in A4, as that is the only smaller number yet to be placed. However, cell B5 states that B4 is also greater than C2. Since there is not another number smaller than 9 that would not have been placed, the scenario is not viable. Therefore, 9 must be placed in A4 and 8 in C2.
19. ) 12, 13, and 14 are the last numbers left to be placed. Cell C2 states that Column B and Row 2 have the same sum. Currently, Column B sums to 32 and Row 2 sums to 33, which means that B4 must be greater than D2 by one. The possible solutions are that B4 is 13 and D2 is 12 (thereby leaving A5 to be 14), or that B4 is 14 and D2 is 13 (leaving A5 to be 12). Since cell C3 states that A5 is greater than B4, that means that A5 cannot have the smallest remaining number, making the former the correct solution. This means that 12 is in D2, 13 is in B4, and 14 is in A5.