Place the numbers 1 through 25 in the grid below:
1 2 3 4 5
A | | | | | |
B | | | | | |
C | | | | | |
D | | | | | |
E | | | | | |
- The sum of each column is odd
- The sum of each row, except C, is even
- The sum of row A is not greater than the sum of any other row
- The sum of the diagonal E1 to A5 is less than that of the diagonal A1 to E5
- A4 + B4 > C4 + D4 + E4
- A1 + B1 = D1 + E1
- A1 > E1
- A1, A3 and B1 are all prime numbers
- (A3 + E3) is a prime number
- A5, D1, D3 and E1 are all squares
- B2, C2 and D2 are ascending consecutive numbers
- B3, C3 and D3 are ascending consecutive numbers
- B5 + D5 = A5 + C5
- (C1)² + (C5)² = (E3)²
- C5 is a two digit number
- D5 is a multiple of E5
- E1 + E3 = E2 + E4 + E5
I agree with Charlie’s answer but preferred to do it manually (or to put it another way, I can’t write programs).
As the sum of A1, B1, D1 and E1 is even (clue 6) and the sum of column 1 is odd (clue 1), C1 must be odd. Possible combinations of C1, C5 and E3 (clue 14) where C1 is odd are:
3,4,5
5,12,13
7,24,25
9,12,15
15,8,17
15,20,25
As C5 is a 2-digit number (clue 15), the 1st and 5th combinations do not work. As E3 is odd (13, 15 or 25) and the sum of E3 and A3 is a prime (clue 9), A3 must be an even number. As E3 is a prime number (clue 8), A3 must be 2 (the only even prime). This is the key to the whole puzzle.
As the sum of A3 and E3 is a prime number (clue 9), the 2nd, 3rd and 6th combinations above do not work. Leaving 9,12,15 as the only possible combination for C1, C5 and E3.
The rest followed easily but is too long-winded to recall here.
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Posted by Farthing
on 2004-12-29 06:14:06 |
Manual Solution...the starting point
