**Question:**

There are 100 switches numbered from 1 to 100. initially in off position.

1. First time I switched on each switch.

2. Second time I switched off switches which has even numbers.

3. Third time I put the switches in opposite position of their state which are numbered with 3 multiplier like 3, 6, 9 and so on.

4. fouth time I put the switches in opposite position of their state which are numbered with 4 multipliers like 4, 8, 12, 16 and so on.

5. I continued this procedure upto 100 times now How many switches are in on position and what are they?

Initially all are off position.

**Step1:** put all switches in on. So all are in on position.

**Step2:** put all even number off switch in off position.

**Step3:** put the switches in opposite position for multipliers of 3 like 3, 6, 9, 12 ...

**Step4:** put the switches in opposite position for multipliers of 4 like 4, 8, 12, 16 ...

** Step5:** put the switches in opposite position for multipliers of 5 like 5, 10, 15, 20....

Now Observe that switches 1,2,3,4 and 5 are will never be changed their positions for remaining future attempts.

Switch 1 and 4 are in on Position.

The factors of 4 are: 1,2 and 4.

The number off factors of 4 is 3.

So we will attempt the 4th switch 3 times(odd number).

So it is opposite to the initial position.

that is why it is in on position.

Switch 2,3,and 5 are in off position.

2,3 and 5 has even number of factors.(for 2 => 1 & 2; for 3 => 1 & 3; for 5 => 1 & 5)

So the number of attempts on these switches are even.

that is why They are in initial position.

All non-perfect squares numbers have even number of factors that is they are in initial position.

All perfect squares numbers have odd number of factors. So switches 1,4,16,25,36,49,64,81 and 100 are in on position.