Here's what I have so far. The 7,13 on the bottom mean that they go in the two open spots in the first column, but I don't know which goes where yet. Same thing for the 6,8 in the bottom row. The oe,ee,oo stuff means the two remaining spots in those rows must be one of those even/odd pairs (not necessarily ordered)
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
And here's how I got there:
Clue 1 means there is always an odd number of odd integers in each column.
Clue 2 means there is always an even number of odd integers in each row, except row C which has an odd number.
Clue 13 and Clue 1 mean that E5 is odd.
Clue 6 and Clue 1 mean that C1 is odd.
Clues 15 and 14 mean C5 = (12, 15, 16, 20, 24). Also, C1 can only be (5, 7, 8, 9, 12, 15, 16, 20) and E3 can only be (13, 15, 17, 20, 25). But since C1 must be odd, we see that C5 = (12, 12, 20, 24), C1 = (5, 7, 9, 15), and E3 = (13, 15, 25, 25). So we see that C5 is even, and E3 is odd.
Clue 9 tells us that of A3 and E3, one is odd and one is even (because we can't have two different numbers from 1-25 add to 2, so they must add to be an ODD prime number). Since we found E3 to be odd, A3 must be even.
Since A3 is even and prime, it must be 2. Then from Clue 8, we know that A1 and B1 must be ODD primes.
From Clue 9, knowing that A3 = 2, and E3 = (13, 15, 25), we see that A3 + E3 = (15, 17, 27). Only 17 is prime. Therefore E3 = 15. This means that C1 = 9 and C5 = 12.
Note: There are 5 squares: 1, 4, 9, 16, 25.
Clue 12 and Clue 10 tell us that D3 is a square with two numbers less than it. So D3 cannot be 1. Also, since A2 is 2, we know that D3 cannot be 4 (if D3 were 4, then B3 would be 2). It also can't be 9 since C1 is. So D3 can only be 16 or 25. But wait, let's remember Clue 12. If D3=16 then C3=15. But E3=15, so D3 can't be 16. Therefore D3 = 25. Which means C3 = 24 and B3 = 23.
Clue 6 says A1 + B1 = D1 + E1. Since we know that A1 and B1 are both ODD primes, their sum must be even. This means that D1 and E1 must both be even or both be odd. Also remember Clue 10 which says they are squares. Remembering that D1 and E1 can't be 25 or 9 (since D3 and C1 are) this leaves only one possible sum: 20. So D1 and E1 are 4 and 16. Since A1 and B1 are both odd primes, there are only a couple possibilities for what they could be. The only ways A1+B1 = 20 is if they were 3 and 17, or 7 and 13.
Clue 10 says that A5 is prime. Well, 4 and 16 are taken on D1 and E1. D3 = 25, and C1 = 9. Therefore A5 = 1, the only remaining square.
From Clue 11 and based on the numbers that are taken already, B2 can only be (5,6,17,18,19,20).
Clue 16 means that E5 must be less than 12 (in order for the next multiple to be less than 25) and odd. Based on the numbers that are taken already, E5 = (3, 5, 7, 11) and D5 = (6/18/21, 10/20, 14/21, 22).
Clue 13 says B5 + D5 = A5 + C5 = 13. Therefore D5 must be less than 13. The only possible cases are where D5 = (6,10) in which cases E5 = (3,5) and B5 = (7, 3).
In either case, 3 is taken. Remember we said A1 and B1 can only be 3 and 17, or 7 and 13. Therefore A1 and B1 must be 7 and 13. But now 7 is taken, which means B5 can only be 3, making E5 = 5, and D5 = 10.
Since 7 is now taken, B2 can only be (17,18,19,20).
Clue 7 says A1 > E1. E1 is either 4 or 16. The only way A1 can be greater than E1 is if E1 = 4.
Clue 17 says E1 + E3 = E2 + E4 + E5. We know E1 + E3 = 19. We also know that E5 = 5. This means that E2 + E4 = 14. Based on the numbers that are left, E2 and E4 can only be 6 and 8.
Clue 4 says the sum of the diagonal E1 to A5 is less than that of the diagonal A1 to E5. Also remember that D2 = B2 + 2. So that means
E1 + D2 + C3 + B4 + A5 < A1 + B2 + C3 + D4 + E5
E1 + D2 + B4 + A5 < A1 + B2 + D4 + E5
4 + (B2+2) + B4 + 1 < A1 + B2 + D4 + 5
B4 + 2 < A1 + D4
Remember that A1 = 7 or 13. Also remember that in order for the individual sums of rows B and D to be even, each one needs exactly one more even and one more odd number. This means that if B4 is even, B2 must be odd. This forces D2 to be odd, which forces D4 to be even. So B4 and D4 must be of the same parity. Hmmm, there are many possible configurations to weed though. I think I'll take a break and post what I have so far.
Edited on December 29, 2004, 2:22 pm
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Posted by nikki
on 2004-12-29 14:18:53 |
Partial Solution - longwinded, no computer
