There are 100 bulbs which are initially in on state.Then the their states are changed in an order mentioned below, 1.Multiples of 2 are toggled.
2.Multiples of 3 are toggled.
3.multiples of 4 are toggled.
.
.
.
.
98. Multiples of 99 are toggled.
99. Multiple of 100 is toggled.
Now which all bulbs are turned on at the end.

• ans for this is the perfect square No.s(1,4,9,16,25,36,49,64,81,100)since these No.s have odd number of factors i.e 1=1*1;4=1*4,2*2 and so on
• 13 years agoHelpfull: Yes(35) No(27)
• 100
  factors of 100= 2,4,5,10,20,25,50,100
  if these bulbs are switched off and on subsequently then 2 is off state, 4 is on , 5 is off,10 on, 20, off, 25 on, 60 off ,100 on,,, last bulb to turn on is 100..
  similarly all perfect square nos are turned on at end..
• 9 years agoHelpfull: Yes(0) No(0)
• perfect squares less than 100 will be turned on at the end .
• 9 years agoHelpfull: Yes(0) No(0)
• only the perfect square numbers will be left turned on, ie; 1,4,9,16,25,36,49,64,81,100
  So, the required answer is 10.
• 8 years agoHelpfull: Yes(0) No(0)
• 10
  no of perfect squares
• 7 years agoHelpfull: Yes(0) No(0)
• 4,9,12,16,25,49,81,100
• 6 years agoHelpfull: Yes(0) No(0)
• All perfect square between 1 and 100
• 6 years agoHelpfull: Yes(0) No(0)