abhelaksh

• Jul 20th, 2005

 

switches dont glow therefore ans is 0...lol

Sandy

• Sep 27th, 2005

 

glowing bulbs..?? doesnt make sense so the answer is 0

sridevi

• Nov 26th, 2005

 

Question: There are 100 bulbs and 100 switches.Each switch is numbered from 1 to 100 and each switch corresponds to each bulb.
 (a)all switches are switched on.
 (b)the switch nnumberes which are divisible by 2 are marked and those switches which are on are put off and which are off are put on.
 (c)the switch  numbers which are divisible by 3 are marked and those switches which are on are put off and which are off are put on.
 (d)this process continued till the number 100.
By the end how many switches are glowing.?

Answer: 10(1,4,9,16,25,36,49,64,81,100)

Dev

• Sep 12th, 2006

 

I think we can solve this problem this way.

I you take a few examples (any number) we will find that it is switched on/off only odd number of times(excluding dividing by 1 and including dividing the number by itself).

Thus as the initial condition is ON, so we have all of them switched OFF; except for the perfect squares,as they have ONE MORE NUMBER to divide them

ie the Suare Root of that number. so they are switched off/on even number of times. So  10(1,4,9,16,25,36,49,64,81 and 100) remains switched ON.

dinesh

• Oct 17th, 2006

 

ANS    "0"   (even the question is wrong , switches cant glow. but if q. was right probably that would still be the right ans.)

first of all , read the question properly..

all 100 switches are switched on.

then switches 2,4,6....100 are "MARKED"( but not switched "OFF") , then all the bulbs which are glowing are switched off, i.e " ALL THE 100 BULBS WHICH ARE GLOWING ARE SWITCHED OFF".

in the next step again all that are switched off are switched on , THUS AT THE END  ALL 100 BULBS ARE SWITCHED "OFF" , coz bulbs remain off for even numbers and on for odd numbers..

Sreedevi Pidaparthi

 Profile Answers by Sreedevi Pidaparthi 

• Jun 28th, 2007

 

If You read the question carefully, it states that initially all the switches are switched on.

then the switch #s divisble by 2 are marked and the switches which are put on are put off.....

so all switches are put off......

now when switch #s divisble by 3 are marked then all switches are put on......

Going by this sequence when we reach 100 all the switches are put off......

So "0" switches are put on should be the correct answer

sathin

 Profile Answers by sathin 

• Aug 25th, 2009

 

Yes the answer is '0'
Regarding statement B  It is said that the switch numbers are divisible
by 2 are marked and those switches which are on are put off and which are
off are put on, we get that all the switches are off from the above step.

For every marking of even number all the switches get off
so as 100 is an even number all the switches get off
so 0 bulbs glow (switches glow may not be technically correct).

ajithg4u

 Profile Answers by ajithg4u 

• Aug 26th, 2009

 

Switch numbers which are perfect squares will be on.
This is because perfect squares have odd number of factors.
eg: 9 factors are 1,3,9
25 factors are 1,5,25

dslamulya

 Profile Answers by dslamulya 

• Mar 28th, 2010

 

This is a logical one and the answer might be all thhe perfect squares from 1 to 100.... 1,4,9,16,25,36,49,64,81,100

jeet_15

 Profile Answers by jeet_15 

• Sep 9th, 2010

 

49 bulbs are glowing.
Initially all bulbs are glowing. then the bulb numbers which are divisible by 2 are switched off. i.e. 2,4,6,....100.
Now the bulb numbers which are divisible by 3 are switched off and which are currently switched off are put on so 3,9,15,21.....99 which are 17 bulbs are switched off too.
and the bulbs 6,12,18,24.....96 which are 16 bulbs are again put on.
so the number of glowing bulbs are = 50+16-17 = 49 simply.

I don't know how others calculated 10 glowing bulbs?

sandeep.thadka

 Profile Answers by sandeep.thadka 

• Jan 18th, 2011

 

Simple logic question:
49 will glow.

Multiples of 2 among 100 numbers          = 50
multiples of 3 among 100 numbers         = 33
multiples of 6(LCM of 2 and 3) among 100 numbers = 16

So first time 50 will  on and next among 50 which are on 17(that is (33-16)) will off and 16(multiples of 6 will now on as previously they were off)

So totalon are 33+16 =49.