1 2 3 4 5
A | | | | | |
B | | | | | |
C | | | | | |
D | | | | | |
E | | | | | |

1. The sum of each column is odd
   
2. The sum of each row, except C, is even
   
3. The sum of row A is not greater than the sum of any other row
   
4. The sum of the diagonal E1 to A5 is less than that of the diagonal A1 to E5
   
5. A4 + B4 > C4 + D4 + E4
   
6. A1 + B1 = D1 + E1
   
7. A1 > E1
   
8. A1, A3 and B1 are all prime numbers
   
9. (A3 + E3) is a prime number
   
10. A5, D1, D3 and E1 are all squares
    
11. B2, C2 and D2 are ascending consecutive numbers
    
12. B3, C3 and D3 are ascending consecutive numbers
    
13. B5 + D5 = A5 + C5
    
14. (C1)² + (C5)² = (E3)²
    
15. C5 is a two digit number
    
16. D5 is a multiple of E5
    
17. E1 + E3 = E2 + E4 + E5

Please visit my previous post for the history, but this is how far I had gotten in it so far:

   1  2  3  4  5 
A |  |  | 2|  | 1|ee,oo
B |  |  |23|  | 3|oe
C | 9|  |24|  |12|oo,ee
D |16|  |25|  |10|oe
E | 4|  |15|  | 5|6,8

  7,13

Ok, I think I got my second wind. We know the full sum of the last row, and the partial sums of the other rows. Let’s see what they are so far. The partial sum of row A is either 10 or 16, and the partial sum of row B is either 39 or 33. The partial sum for row C is 45, and the partial sum for row D is 51. Row E’s sum is 38. From Clue 3 we know that row A will have to be less than 38. From row B’s partial sum of 33 you might think the total sum could be less than 38, but with two slots left to fill and the two smallest numbers left to be placed are 11 and 14, row B’s minimum would be 58.

So row A must be less than or equal to 38. The two smallest available numbers are 11 and 14. However, A2 and A4 must both be even or both be odd. So we’ll have to ditch one of them, but let’s start with those anyway and see how much wiggle room we have. Also remember we don’t know if A1 is 7 or 13 yet. So let’s say row A = 13 + 11 + 2 + 14 + 1 = 41. Woah, that’s too big. So A1 can’t be 13. Let’s try row A = 7 + 11 + 2 + 14 + 1 = 35. Ok we have a wiggle room of three. The next smallest available number is 17, which is odd meaning we’ll have to ditch the 14 so that 11 and 17 can be odd together. Then we get row A = 7 + 11 + 2 + 17 + 1 = 38. Well, this is legal, but we don’t have anymore wiggle room, so this must be the answer for row A.

Remembering Clue 1, in order for the sum of column 2 to be odd, the three empty spots (B2 – D2) need to have no odd numbers or exactly two odd numbers since we know A2 is odd and E2 is even. Well, since Clue 11 says that B2 – D2 are ascending consecutive numbers, we know there will be at least one more odd number. So having no more odd numbers is impossible. Therefore we must have two more odd numbers. And the only way for that to happen is if B2 is odd, C2 is even, and D2 is odd.

Since B2 is odd, and in my previous post I said that B2 could be (17, 18, 19, 20), I know even more about B2. 17 is taken by row A, so now the only odd number left is 19. Therefore B2 = 19, C2 = 20, and D2 = 21.

The only numbers left for B4, C4 and D4 now are 14, 18, and 22, so they’ll all be even which is consistent with what we need for our column and row sums according to Clues 1 and 2.

Clue 5 says A4 + B4 > C4 + D4 + E4. Well, C4 and D4 must be two of 14,18,22. We’ll start by saying E4 is 6 (and if we have enough wiggle room in the end, it could be 8). So the three sums of C4 + D4 + E4 are 46 (when B4 = 14), 42 (when B4 = 18) and 38 (when B4 = 22).

Remember A4 can only be 11 or 17. So in the first case, where B4 = 14, neither 14+11 nor 14+17 is greater than 46, so this is not a possibility. In the second case, where B4 = 18, neither 18+11 nor 18+17 is greater than 42, so this is not a possibility. Therefore B4 must be 22.

In my previous post I used Clue 4 and found that B4 + 2 < A1 + D4. Knowing that B4=22 and A1=7, we have D4 > 17. The only possible numbers for D4 now are 14 and 18. Therefore D4 = 18, which leaves C4 = 14.

Ok, back to Clue 5.
A4 + B4 > C4 + D4 + E4
A4 + 22 > 14 + 18 + E4
A4 > 10 + E4

If E4 can only be 6 or 8, and A4 can only be 11 or 17. This means that if E4 = 8, A4 > 18. This is not possible. So E4 = 6 and A4 > 16. This is only possible if A4 = 17.

Haha! I have conquered the puzzle! I have to say, I think this was harder than a level 3 difficulty, but that’s just me.

   1  2  3  4  5 
A | 7|11| 2|17| 1|
B |13|19|23|22| 3|
C | 9|20|24|14|12|
D |16|21|25|18|10|
E | 4| 8|15| 6| 5|