I am not convinced of the asnwer for this one.. There are

Question

I am not convinced of the asnwer for this one..

There are 1000 rooms in a row along a long corridor. Initially the first room contains 1000 people and the remaining rooms are empty. Each minute, the following happens: for each room containing more than one person, someone in that room decides it is too crowded and moves to the next room. All these movements are simultaneous (so nobody moves more than once within a minute). After one hour, how many different rooms will have people in them?

twixt · Answer

From scratch after 10 seconds I would say 58.. details if i am not totally wrong

PracticeMore · Answer

From the 1st room if every min a person comes out, it will be 60 of them. Now if each continue to move once they double then the time will go this way

1st min
Room 2 : a person came in 1st min. Then 2nd joined in 2nd min. And then one of them left in 3rd min. While in the same 3rd min a new member can join out of 998 people. So room strength every minute will remain 2 , and so every minute someone must leave. 

So 2nd min onwards, the number on room 2 becomes const, ie one joins and one leaves. 

Similarly room 3 will have one person added and then to increase to 2, it will need another 1 min.

So every room that gets filled takes a minute for one person to get in.

So in 60 minutes, the room gets filled every odd minutes, ie 1, 3, 5, and so on. so last minute when the room will be filled will be 59 th min

tn = 59= 1 + (n-1)2 , so n = 30

So number of rooms that will be occupied in an hour will be 30 plus the one they already were in , ie 31.


Let us know the OA

Dan · Answer

PracticeMore,

"1st min 
Room 2 : a person came in 1st min. Then 2nd joined in 2nd min. And then one of them left in 3rd min. While in the same 3rd min a new member can join out of 998 people. So room strength every minute will remain 2 , and so every minute someone must leave."

If the 2nd joined in the 2nd min., it becomes a crowd in the 2nd min., and the person who came in the 1st min. will leave in the 2nd min., and not in the 3rd min.

carsen · Answer

Good one here. My reasoning...

Minutes ---> Number of people/room

0 ---> 1000 
1 ---> 999 - 1 (First person moves to room 2)
2 ---> 998 - 2 (Then from room 1 to room 2)
3 ---> 997 - 2 - 1
4 ---> 996 - 2 - 2 (2 rooms with 2people in it)
5 ---> 995 - 2 - 2 - 1
.
.
.
.
10 ---> 990 -2-2-2-2-2 (5 rooms with 2 people in it)
.
.
.
.
60 ---> 940 - 2 - 2 - 2 - 2 - 2....2 (30 rooms with 2 people in it)

If we see the above pattern, we know that, at a time there can only be maximum 2 people in a room. For every odd, the sequence ends with one, and for even the sequence ends with 2. (Tip : for every even number out of 1000, the number of room is half of it. example ..After 20 minutes, 20 people come out, and the sequence will be 980+ 10 rooms of 2 people.

Hence, after 60 minutes, it wud be 60/2=30. 2 peoplein 30 rooms and 1 room with 940 people. Total of 31 rooms. 
Let me know if this is wrong.

Answer  31

Dan · Answer

Carsen, you're making the same assumption that the first person who moved to room 2 in the first minute will not move until minute 3. Why not in minute 2?

carsen · Answer

Hello Dan

It does not matter if it is the 1st person who moves to the third room or the 2nd person moving to the 3rd room. The head count in the room at a given time will never be more than 2. Every time there are 2, then one person moves to the other room, like a string of beads.

I hope this helps.

Dan · Answer

yes thanks...good explanations Carsen and PracticeMore. Two different ways. OA 31. Here is a 3rd one.

"We can prove by induction on n that the following pattern holds for 0 <= n <= 499: after 2n minutes, the first room contains 1000 - 2n people and the next n rooms each contain 2 people, and after 2n + 1 minutes, the first room contains 1000 - (2n + 1) people, the next n rooms each contain 2 people, and the next room after that contains 1 person. So, after 60 minutes, we have one room with 940 people and 30 rooms with 2 people each."

I am not convinced of the asnwer for this one.. There are

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