Step 1: A1 = 1. Clue A1 reveals that E5 = 25 since that is the only location for it.

Step 2: Clue E5 (and Clue A1) reveals B2 = 4, C3 = 9, and D4 = 16. Additional clues are revealed. Notice that the top left 2×2 grid must contain numbers 1-4. the top left 3×3 grid contains 1-9, and the top left 4×4 grid contains 1-16. Hence, it’s easiest to start in the top left region, where B1 and A2 must be 2 and 3 in some order.

Step 3: Clue C3 indicates that B1 = 3 and thus A2 = 2, since the top row has no even numbers.

Aside: Number 5 is in either C1 or A3.

Step 4: For the group of numbers 10-16, the position of number 10 can be deducted to be in A4, since there are no evens in the top row by Clue C3. A4 = 10.

Step 5: Now, the other half of Clue C3 (no odd numbers adjacent to 10) helps with many other cells. Firstly, A3 = 6 and B3 = 8, because they are the only evens in the 3×3 square. Therefore, C1 = 5 and C2 = 7.

Also, since 11 and 17 can’t be adjacent to 10, D1 = 11 and E1 = 17.

Step 6: There are already 2 primes in Column C (numbers 5 and 7), so by Clue D4, column D needs at least 3 primes. This means that number 13 has to go in either D2 or D3, and D5 contains another prime. 19 and 23 are the only primes that could fit, but 19 in D5 is impossible given the increasing order restriction, so D5 = 23. Hence, E4 = 24.

Step 7: Clue B2 requires an odd number in row 4 and it can only be 15, since 13 is in D2 or D3. Due to increasing order, C4 = 15.

Step 8: A5 and B5 are both evens (adjacent to 10), and B5 can’t contain 22, leaving a gap to 23 in D5. Hence, A5 = 18, B5 = 20 and E2 = 19.

Step 9: Column C still needs an even number, and it must be C5 = 22. So, E3 = 21.

Step 10: The final three numbers to place are 12, 13 and 14 into D2, D3 and B4 in some order, with 13 in column D. Clue D4 says that row 4 has the fewest primes. Currently, neither row 3 and row 4 have any primes, so 13 has to go row 3 to make row 4 have strictly fewer. D3 = 13, and by increasing order: D2 = 12 and B4 = 14.