There are 100 bulbs in a room. 100 strangers have been accumulated in the adjacent room. The first one goes and lights up every bulb. The second one goes and switches off all the even numbered bulbs – second, fourth, sixth... and so on. The third one goes and reverses the current position of every third bulb (third, sixth, ninth… and so on.) i.e. if the bulb is lit, he switches it off and if the bulb is off, he switches it on. All the 100 strangers progresses in the similar fashion.
After the last person has done what he wanted, which bulbs will be lit and which ones will be switched off ?

• all the perfect square will be in on state, so for 100 bulbs , 10 bulbs will be in on state(i.e, perfect squares are 1,4,9,16,25,36,49,64,81,100. so totally 10 bulbs) and remaining bulbs are in off state.
• 8 years agoHelpfull: Yes(3) No(0)
• First think who will operate each bulb, obviously person #2 will do all the even numbers, and say person #10 will operate all the bulbs that end in a zero. So who would operate for example bulb 48:
  
  Persons numbered: 1 & 48, 2 & 24, 3 & 16, 4 & 12, 6 & 8 ........
  
  That is all the factors (numbers by which 48 is divisible) will be in pairs. This means that for every person who switches a bulb on there will be someone to switch it off. This willl result in the bulb being back at it's original state.
  
  So why aren't all the bulbs off?
  
  Think of bulb 36:-
  
  The factors are: 1, 2, 3, 6, 9, 12, 18, 36
  OR 1 & 36, 2 & 18, 3 & 12 and 6
  
  Previously the pairs cancelled, if 2 flicks the switch one way then 18 would flick it back, but in this case 6 is not paired, it's by itself, there is no body to cancel out his action. This is true of all the square numbers.
  
  There are 10 square numbers between 1 and 100 (1, 4, 9, 16, 25, 36, 49, 64, 81 & 100) hence 10 bulbs remain on.
• 7 years agoHelpfull: Yes(1) No(0)
• all the perfect square nos lit remaining are off
• 9 years agoHelpfull: Yes(0) No(5)
• Ponder over the bulb number 56, people will visit it for every divisor it has. So 56 has 1 & 56, 2 & 28, 4 & 14, 7 & 8. So on pass 1, the 1st person will light the bulb; pass 2, 2nd one will switch it off; pass 4, light it; pass 7, switch it off; pass 8, light it; pass 14, switch it off; pass 28, light it; pass 56, switch it off.
  For each pair of divisors the bulb will just end up back in its preliminary state. But there are cases in which the pair of divisor has similar number for example bulb number 16. 16 has the divisors 1 & 16, 2 & 8, 4&4. But 4 is recurring because 16 is a perfect square, so you will only visit bulb number 16, on pass 1, 2, 4, 8 and 16… leaving it lit at the end. So only perfect square bulbs will be lit at the end.
• 7 years agoHelpfull: Yes(0) No(0)
• only the perfect square bulbs will be lit at the end
• 4 years agoHelpfull: Yes(0) No(0)