Mathematics - Permutation & Combination Question with Solution | TestHub
D₁, D₂, ..., D₁₀₀₀ are 1000 doors, and P₁, P₂, ..., P₁₀₀₀ are 1000 persons. Initially, all the doors are closed. P₁ opens all the doors. Then, P₂ closes D₂, D₄, D₆, ..., D₉₉₈, D₁₀₀₀. Then P₃ changes the status of D₃, D₆, D₉, D₁₂, etc. (doors having numbers which are multiples of 3). Changing the status of a door means closing it if it is open and opening it if it is closed. Then P₄ changes the status of D₄, D₈, D₁₂, D₁₆, etc. (doors having numbers which are multiples of 4). This continues until P₁₀₀₀ changes the status of D₁₀₀₀.
Finally, how many doors are open?
Options:
Answer:
Solution:
Consider any door, for example, D₇₂. It is operated by P₁, P₂, P₃, P₄, P₆, P₈, P₉, P₁₂, P₁₈, P₂₄, P₃₆, P₇₂. (Remember that Dₘ is operated by Pₙ if is a multiple of .) Here, 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72 are all the factors of 72. Initially, all the doors are closed. Therefore, if an odd number of persons operate it, it will be finally open. Otherwise, it will be finally closed.
Thus, Dₘ will be finally open if has an odd number of factors. We know that has an odd number of factors if and only if is a perfect square.
Therefore, are the numbers of the doors that are finally open.
The number of doors finally open is 31.
