Question

There are 100 switches in a board. 100 person have been seated in the row. The first one goes and turns on every switch. The second one goes
and switches off all the odd numbered switch (switches 1,3,5...). The third one goes and reverses the current position of every third switch
(switches 3,6,9...). The fourth one goes and turns on every fourth and fifth switch (switches 4,8,12... and 5,10,15...). All the remaining person
progresses with what person 2,3 and 4 did in the round robin fashion.
After the last person has done what he wanted, what will be the position of the switches which are in perfect number positions?
A
On, On
B
On, Off
Off, On
С
Off, Off
D​

Answer 1

Answer:

Answer for this question please

Answer 2

The correct answer for the position of the switches which are in perfect number positions is off, off. So option D is the correct answer.

Step-by-step explanation:

There are $100$ bulbs and $100$ men.

The positive integer that is equal to the sum of its all factors is called a perfect number.

So $1$ to $100$, there are $2$ perfect numbers those are $6$ and $28$.

Factors of $6$ are $1,2,3,6$.

That means person1, person2, person3, and person4 operate the 6th switch.

So after $4$ operations, the 6th switch will be in the off position.

Similarly, factors of $28$ is $1,2,4,7,14,28$.

That means person1, person2, person4, person7, person14, person28 operate the 28th switch.

So after6 operations, the 28th switch will be in the off position.

Hence the position of the switches which are in perfect number positions is off, off.