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Tick, Trick, and Track have 20, 23, and 25 tickets, respectively, for the carousel
at the fair in Duckburg. They agree that they will only ride all three
ride together, for which they must each give up one of their tickets.
tickets. In addition, before a ride, they can, if they wish, give tickets to each other as many times as they like according to
redistribute tickets among themselves as often as they like according to the following rule: If one has an
even number of tickets, he can give half of his tickets to any of the other two.
to any of the other two.
Can it happen that after any trip
(a) exactly one has no ticket left,
(b) exactly two have no ticket left,
(c) all tickets are given away?
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Mathematics
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Re: Mathematics
by Guest » Fri Feb 17, 2023 2:34 am
(a) We can show that it is not possible for exactly one of Tick, Trick, and Track to have no tickets left after a ride.
Suppose that after a ride, only Tick has no tickets left. Since each of them gave up one ticket, this means that Tick originally had only one ticket. However, in order for all three of them to ride the carousel, each of them must give up a ticket, which is not possible if Tick only had one ticket to begin with. Therefore, it is impossible for only Tick to have no tickets left after a ride, and similarly for Trick and Track.
(b) We can show that it is possible for exactly two of Tick, Trick, and Track to have no tickets left after a ride.
Suppose that initially Tick, Trick, and Track have 20, 23, and 25 tickets, respectively.
First, Tick gives half of his tickets (10) to Trick, so that Tick has 10 tickets and Trick has 33 tickets. Then, Trick gives half of his tickets (16) to Track, so that Trick has 17 tickets and Track has 41 tickets. Finally, Track gives half of his tickets (20.5, rounded down to 20) to Tick, so that Tick has 30 tickets and Track has 20 tickets.
At this point, Tick and Trick can ride the carousel together, since they have a total of 47 tickets between them, which is more than the 20 tickets required for all three of them to ride. After the ride, Tick has 29 tickets and Trick has 16 tickets, so exactly two of them have no tickets left.
(c) We can also show that it is possible for all the tickets to be given away after a ride.
Suppose that initially Tick, Trick, and Track have 10, 10, and 32 tickets, respectively.
First, Tick gives half of his tickets (5) to Trick, so that Tick has 5 tickets and Trick has 15 tickets. Then, Track gives half of his tickets (16) to Tick, so that Tick has 21 tickets and Track has 16 tickets. Finally, Trick gives half of his tickets (7) to Track, so that Trick has 8 tickets and Track has 23 tickets.
At this point, all three of them can ride the carousel together, since they have a total of 52 tickets between them, which is exactly the number of tickets they started with. After the ride, each of them has given up one ticket, so all the tickets have been given away.
Therefore, it is possible for exactly two of Tick, Trick, and Track to have no tickets left after a ride, and it is possible for all the tickets to be given away after a ride, but it is not possible for exactly one of them to have no tickets left after a ride.
Suppose that after a ride, only Tick has no tickets left. Since each of them gave up one ticket, this means that Tick originally had only one ticket. However, in order for all three of them to ride the carousel, each of them must give up a ticket, which is not possible if Tick only had one ticket to begin with. Therefore, it is impossible for only Tick to have no tickets left after a ride, and similarly for Trick and Track.
(b) We can show that it is possible for exactly two of Tick, Trick, and Track to have no tickets left after a ride.
Suppose that initially Tick, Trick, and Track have 20, 23, and 25 tickets, respectively.
First, Tick gives half of his tickets (10) to Trick, so that Tick has 10 tickets and Trick has 33 tickets. Then, Trick gives half of his tickets (16) to Track, so that Trick has 17 tickets and Track has 41 tickets. Finally, Track gives half of his tickets (20.5, rounded down to 20) to Tick, so that Tick has 30 tickets and Track has 20 tickets.
At this point, Tick and Trick can ride the carousel together, since they have a total of 47 tickets between them, which is more than the 20 tickets required for all three of them to ride. After the ride, Tick has 29 tickets and Trick has 16 tickets, so exactly two of them have no tickets left.
(c) We can also show that it is possible for all the tickets to be given away after a ride.
Suppose that initially Tick, Trick, and Track have 10, 10, and 32 tickets, respectively.
First, Tick gives half of his tickets (5) to Trick, so that Tick has 5 tickets and Trick has 15 tickets. Then, Track gives half of his tickets (16) to Tick, so that Tick has 21 tickets and Track has 16 tickets. Finally, Trick gives half of his tickets (7) to Track, so that Trick has 8 tickets and Track has 23 tickets.
At this point, all three of them can ride the carousel together, since they have a total of 52 tickets between them, which is exactly the number of tickets they started with. After the ride, each of them has given up one ticket, so all the tickets have been given away.
Therefore, it is possible for exactly two of Tick, Trick, and Track to have no tickets left after a ride, and it is possible for all the tickets to be given away after a ride, but it is not possible for exactly one of them to have no tickets left after a ride.
- Guest
Re: Mathematics
by Kokez » Fri Feb 17, 2023 10:41 am
is this answer correct
Ticket_problem (1).pdf- (61.94 KiB) Downloaded 90 times
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