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If D1, D2, D3, ..... D1000 are 1000 doo...

If `D_1, D_2, D_3, ..... D_1000` are 1000 doors and `P_1, P_2, P_3, ..... P_1000` are 1000 persons. Initially all doorsare closed. Changing the status of doors means closing the door if it is open or opening it if it is closed. `P_1` changes the status of all doors. Then `P_2` changes the status of `D_2, D_4, D_6, .... D_1000` (doors having numbers which are multiples of 2). Then `P_3`; changes the status of `D_3, D_6, D_9, .....D_999` (doors having number which are multiples of 3) and this process is continued till `P_1000` changes the status of `D_1000`, then the doors which are finally open is/are

A

`D_(961)`

B

`D_(269)`

C

`D_(413)`

D

`D_(729)`

Text Solution

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The correct Answer is:
To solve the problem of determining which doors remain open after all 1000 persons have changed the status of the doors, we can follow these steps: ### Step-by-Step Solution: 1. **Understanding the Problem**: - We have 1000 doors, all initially closed. - Each person toggles the state of certain doors based on their number. For example, `P1` toggles all doors, `P2` toggles every 2nd door, `P3` toggles every 3rd door, and so on up to `P1000`. 2. **Identifying the Toggling Pattern**: - A door `D_n` will be toggled by every person whose number is a divisor of `n`. For example, door `D6` will be toggled by `P1`, `P2`, `P3`, `P6`. 3. **Counting the Toggles**: - A door will end up being open if it is toggled an odd number of times. This occurs when the number of divisors of the door number is odd. 4. **Understanding Divisors**: - A number has an odd number of divisors if and only if it is a perfect square. This is because divisors generally come in pairs (e.g., for `12`, the pairs are `(1, 12)`, `(2, 6)`, `(3, 4)`), but a perfect square has one unpaired divisor (e.g., for `9`, the pairs are `(1, 9)`, `(3, 3)`). 5. **Finding Perfect Squares**: - We need to find all perfect squares between `1` and `1000`. The perfect squares are `1^2, 2^2, 3^2, ..., n^2` where `n^2 ≤ 1000`. - The largest integer `n` such that `n^2 ≤ 1000` is `31` because `31^2 = 961` and `32^2 = 1024` which exceeds `1000`. 6. **Listing the Open Doors**: - The perfect squares from `1` to `31` are: - `1^2 = 1` - `2^2 = 4` - `3^2 = 9` - `4^2 = 16` - `5^2 = 25` - `6^2 = 36` - `7^2 = 49` - `8^2 = 64` - `9^2 = 81` - `10^2 = 100` - `11^2 = 121` - `12^2 = 144` - `13^2 = 169` - `14^2 = 196` - `15^2 = 225` - `16^2 = 256` - `17^2 = 289` - `18^2 = 324` - `19^2 = 361` - `20^2 = 400` - `21^2 = 441` - `22^2 = 484` - `23^2 = 529` - `24^2 = 576` - `25^2 = 625` - `26^2 = 676` - `27^2 = 729` - `28^2 = 784` - `29^2 = 841` - `30^2 = 900` - `31^2 = 961` ### Conclusion: The doors that remain open are the doors numbered: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961.

To solve the problem of determining which doors remain open after all 1000 persons have changed the status of the doors, we can follow these steps: ### Step-by-Step Solution: 1. **Understanding the Problem**: - We have 1000 doors, all initially closed. - Each person toggles the state of certain doors based on their number. For example, `P1` toggles all doors, `P2` toggles every 2nd door, `P3` toggles every 3rd door, and so on up to `P1000`. ...
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