**Solution:**

This is a little tricky. Say switches nos are 1 to 100.

Here

after 1st round, all switches are ON (All were OFF, so toggling will make all ON)

after 2nd round, all switches divisible by 2 (switch no divisible by 2) are TOGGLED

after 3rd round, all switches divisible by 3 (switch no divisible by 3) are TOGGLED

after 4th round, all switches divisible by 4 (switch no divisible by 4) are TOGGLED

.......

Here

after 1st round, all switches are ON (All were OFF, so toggling will make all ON)

after 2nd round, all switches divisible by 2 (switch no divisible by 2) are TOGGLED

after 3rd round, all switches divisible by 3 (switch no divisible by 3) are TOGGLED

after 4th round, all switches divisible by 4 (switch no divisible by 4) are TOGGLED

.......

In other way, we can see that all switches having ODD no of divisors are going on be ON, and all switches with EVEN no of Divisors are going to be OFF.

e.g. Switch No

1 == >> divisor is 1 (ODD count), will be ON

2 == >> divisors are 1, 2 (EVEN count), will be OFF

3 == >> divisors are 1,3 (EVEN count), will be OFF

4 == >> divisors are 1, 2, 4 (ODD count), will be ON

Now how to find out all nos from 1 to 100 having ODD no of divisors. This is again tricky. Only SQUARE Nos will have ODD no of divisors :-) so simple.

So nos with ODD divisor count are: 1,4,9,16,25,36,49,64,81,100

So nos with ODD divisor count are: 1,4,9,16,25,36,49,64,81,100

So answer is: Switches with SQUARED no will be ON after 100 rounds (All other will be OFF).

**POST YOUR OPINION IF YOU HAVE BETTER SOLUTION**

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