A big hostel, there are 1,000 rooms. in that hostel only even numbers are used for room numbers, i.e. the room numbers are 2, 4, 6, ...., 1998, 2000. all the rooms have one resident each. one fine morning, the warden calls all the residents and tells them to go back to their rooms as well as multiples of their room numbers. when a guy visits a room and finds the door open, he closes it, and if the door is closed, he opens it, all 1,000 guys do this operation. all the doors were open initially.
Answers
Answer:
Step-by-step explanation:
The residents are asked to go to their rooms as well as multiples of their room numbers to open or close the doors.
Room Number -------- Resident of room number visiting
2 2(close)
4 2(close), 4(open)
6 2(close), 6(open)
8 2(close), 4(open), 8(close) and so on...
We see that numbers with odd number of even factors will have door closed.
Since each room number is even, so it can be written as 2^p× b^q × c^r
where b and c are prime numbers.
Total number of factors = (p + 1) (q + 1) (r + 1) .... (1)
and total number of even factors = p(q + 1) (r + 1).....(2)
(31) Our aim is to make (2) odd to obtain the room numbers whose door is closed.
To make (2) odd, q should be even, r should be even and p should be odd.
Also, the total number of rooms in the hotel are 1000(numbered 2, 4,6, ..., 2000).
We know that the perfect squares have odd factors, so they will be closed. So, the last room that is closed will be the room which is nearest to room number 2000 and is a perfect square.
Since 961 is a perfect square closest to 1000 (because total number of rooms are 1000).
∴ room number 2 × 961 i.e. 1922 will be the last room that is closed.
(32) The closed rooms between 2 and 76 are:
2(1 is a perfect square),
8 (4 is a perfect square), 18 (9 is a perfect square), 32 (16 is a perfect square),
50 (25 is a perfect square),
72 (36 is a perfect square).
Since, next closed room will be 98
(98/2 = 49 is a perfect square).
Now, 6 rooms are closed from 2 to 76, so we have 12 numbers (2 for each room) for these 6 rooms.
∴ 38th open room will be 76 + 12 = 88
(33) If only 500 guys are asked to do the task, then room number 1002 to 2000 will have and one factor less and that is itself.
So, previously closed rooms will open and open rooms will be closed.
Since 2000/2 is not a perfect square. It was open when all were visiting, so this time it will be closed.
Hence Room number 2000 will be the last room which is closed.
(34) Now, number of perfect squares in between 1 to 500 are 22
[Numbered from 1 to 22【(22)^2 = 441】]
and the number of perfect squares between 501 to 1000 are 9 [from 23 to 31].
Total number of closed rooms in last question = Number of closed rooms from 1 to 500 + Number of closed room from 501 to 1000
= 22 + (500 – 9)
= 513
(35) A lazy person would like to stay in the room having number more than 1000 as he has to visit only once and that is in his own room.
HOPE you understood the solution