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
(In reply to
solution by alex)
not sure if i'm suppose to show my work but...
since most of the hints narrow down the possible numbers belonging in certain squares, i tried to find all the different possibilities and cross reference them with the possibilities of connected hints. for example...
a set of 3 consecutive numbers whose final member is a square can only be {2,3,4},{7,8,9},{14,15,16}, and {23,24,25} - hints 10 and 12)
if A3 (which must be an odd number because if A1 + B1 = D1 + E1, and the sum of each column is an odd number, then C1 must be an odd number) + C5 (which is a 2 digit number) = E3 (which, logically, must also be a 2 digit number, and when added to a prime number the sum is another prime number) then you're left with 4 possibilities, which can be reduced to one if you bear in mind that there are only five squares and four of these have assigned positions. - hints 1,6, 8, 9, 10, 14, 15.
etc.
that all sounds really confusing, and it is. i'm sure there was an easy way to go about this, but that's how i did it.
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Posted by alex
on 2005-01-14 00:03:34 |
re: solution
