Challenging question

There are 100 bulbs and each having separate switches and on at the initial stage.

First man enters and switches off the bulbs, which are the multiples of 2.

Next man enters and switches on the bulbs which are off and switches off the bulbs which are on, which are multiples of 3.

Next man does the same as above for the multiples of 4.

Next man does it for the multiples of 5.

This process is repeated up to the multiples of 100.

So finally how many bulbs will be glowing

plz explain

Nice.
The bulbs glowing will be equal to the amount of numbers not divisible by another number and 1. and also one extra bulb, the first one. Actually, the third man, who switches off the multiples of 4 has no work to do, so he can just go home.
The men who have to switch off bulbs who have prime numbers are the only ones doing any work.
so, the 2,3,5,7,11,13 guys are all working, while the 4th, 6th, 8th and 10th all have no work to do?
Also the guys who have to switch off bulbs with numbers greater than 50 only have to flip one switch at most.

This is a mechanical version of the sieve of Eratosthenes.

Since there are 25 primes numbers less than 100, and the first bulb remains lit,
then there are 26 bulbs glowing.

Here is the list:

"There are 25 prime numbers between 1 and 100. They are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97."

Now, the explanation is thus: any bulb that has a composite number gets switched off, while any bulb that has a prime number stays on, while the first bulb always remains on. So the number of bulbs that are on, is equivalent to the number of prime numbers below 100 and in addition, the first bulb remains on, so the number of bulbs on is one more than the number of prime numbers.

OK!
Trick Question:
The last guy who is guy 100,
switches all the bulbs that were off On, turns off bulb 100,