Playing soccer with 3 goes as follows: 2 field players try to make a goal past the goalkeeper, the one who makes the goal stands goalman for next game, etc. Arne, Bart and Cauchy played this game. Later, they tell their math teacher that A stood 12 times on the field, B 21 times on the field, C 8 times in the goal. Their teacher knows who made the 6th goal. Who made it?
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Lucky707
09.08.2004 21:05
My attempt...
Let's let A and C be represented by D
Clearly, B cannot stand in the goal two straight times. B also stood in the goal one time more than D. Therefore, B must have started in goal and ended in goal as such...
BDBDBDBDBDBDBDB....etc.... B
If B started in goal, D scored the first goal, B scored the second, etc... meaning D (A or C) scored all the odd numbered goals and B scored all the even numbered goals...
Therefore, B must've scored the 6th goal
Peter
09.08.2004 21:31
eum.. yes.. but how do you proof there's nowhere ...BDDB... in the list? strings ..BCAB.. and ..BACB.. are not contradicting.. So you're still missing something...
Peter
18.08.2004 11:17
Hey lucky you were almost there, why don't you continue? You've found the hardest part yet, but you miss the easy link to make it consistent
Peter
25.09.2005 03:13
No one up for completing?
chess64
25.09.2005 03:18
Wow, this topic is old!
Arne
25.09.2005 12:17
I didn't know I am that good at soccer
Kalle
25.09.2005 13:26
We know that A and C were in the goal 21 times. C was there 8 times, so A was goalie 13 times. Similar reasoning, B was goalie 4 times. So in total there were 25 rounds played, and C was on the field 17 times. The only way for A to be the goalie 13 times in 25 games is if he starts out being goalie, and then is goalie as often as possible. This would mean he starts out being the goalie, then is it the third round, the fifth, the seventh.... The player who was goalie the seventh round was the one who made the sixth goal. So A made the sixth goal.
Peter
27.09.2005 15:20
That's correct, Kalle.