I am no further than I was when I started reading it !!!



Quote by celticq
Okay - assuming
(1) 1,2,3,4 > 5, 6,7 ,8 Then 9,10,11,12 are all = G
(2) a 1,6,7,8 > 5,10,11,12 Then F = 1,5 or 9 (not 9, you've already identified it as good)]
b 1,6,7,8 = 5,10,11,12 Then F = 2, 3 or 4
c 1,6,7,8 < 5,10,11,12 Then F = 6,7or 8
Right I've got this far now - I can now get the fake from result a
but not from result b or c
Quote by celticq
Okay by substituting the condoms for 11 tealights and a stapler (which I have pretended looks like a tealight) I think I have now got it.
If fake is 2, 3 or 4 then its heavier
If its 6, 7 or 8 then its lighter
Hurray! Either way one weigh decides the fake.
Quote by cplonamission
OK, this better be right!
Number the Condoms 1 to 12.
Weigh 1-4 against 5-8 - if the same go to A if different go to B
A - get a condom you know is not fake (IE 1-8 - call it 0) weigh 0 and 9 against 10 and 11
-if the same go to (1) if 0and 9 heavier go to (2) if 10 and 11 heavier go to (3)
B - ok condoms 9-12 are real. Take one real one ( called 0) weigh 0, 6 and 3 against 5,7 and 2.
- if same go to (4). If 0,6,3 heavier go to (5) if 5,7,2 heavier go to (6)
(1) - we know 12 is the fake and weigh against any other condom. Yes.
(2) - weigh 10 against 11. If balanced 9 is fake and heavy, Yes. if 10 is heavier then that is the fake. No. According to condition A, (0,9) is heavier than (10,11), therefore if 10 and 11 don't balance, the fake is the lighter of those two.
(3) - weigh 10 against 11, if balanced 9 is fake and light, if one is heavier that is the
(4) Weigh 1 against 4, if the same 8 is fake and light. If one if heavier it is the is only true if on your first weighing (1,2,3,4) is heavier than (5,6,7,8). If number 1 was fake and light, you would not get the right answer, you would get the answer that number 4 was fake and heavy.
(5) Weigh 5 against 7, if the same 3 is fake and Number 2 could be fake and light.. If one is lighter then it is
(6) Either 2 is heavy or 6 is light, weight one and Number 3 could be light.
Nor that better be right after I typed it, I am sure there is a simpler way by giving all the condoms a 3 digit no and having weighting combinations but couldn't be arsed to figure it out.
Quote by Ice Pie:bounce:
CQ has actually solved it
Quote by Ice Pie
I think
Quote by Ice Pie
OK, this better be right!
Number the Condoms 1 to 12.
Weigh 1-4 against 5-8 - if the same go to A if different go to B
A - get a condom you know is not fake (IE 1-8 - call it 0) weigh 0 and 9 against 10 and 11
-if the same go to (1) if 0and 9 heavier go to (2) if 10 and 11 heavier go to (3)
B - ok condoms 9-12 are real. Take one real one ( called 0) weigh 0, 6 and 3 against 5,7 and 2.
- if same go to (4). If 0,6,3 heavier go to (5) if 5,7,2 heavier go to (6)
(1) - we know 12 is the fake and weigh against any other condom. Yes.
(2) - weigh 10 against 11. If balanced 9 is fake and heavy, Yes. if 10 is heavier then that is the fake. No. According to condition A, (0,9) is heavier than (10,11), therefore if 10 and 11 don't balance, the fake is the lighter of those two.
(3) - weigh 10 against 11, if balanced 9 is fake and light, if one is heavier that is the
(4) Weigh 1 against 4, if the same 8 is fake and light. If one if heavier it is the is only true if on your first weighing (1,2,3,4) is heavier than (5,6,7,8). If number 1 was fake and light, you would not get the right answer, you would get the answer that number 4 was fake and heavy.
(5) Weigh 5 against 7, if the same 3 is fake and Number 2 could be fake and light.. If one is lighter then it is
(6) Either 2 is heavy or 6 is light, weight one and Number 3 could be light.
Nor that better be right after I typed it, I am sure there is a simpler way by giving all the condoms a 3 digit no and having weighting combinations but couldn't be arsed to figure it out.