If U can.....answer this

[vssoma]

Question: 100 bulbs

There are 100 light bulbs lined up in a row in a long room. Each bulb has its own switch and is currently switched off. The room has an entry door and an exit door. There are 100 people lined up outside the entry door. Each bulb is numbered consecutively from 1 to 100. So is each person. Person No. 1 enters the room, switches on every bulb, and exits. Person No. 2 enters and flips the switch on every second bulb (turning off bulbs 2, 4, 6, ). Person No. 3 enters and flips the switch on every third bulb (changing the state on bulbs 3, 6, 9, ). This continues until all 100 people have passed through the room.

What is the final state of bulb No. 64? And how many of the light bulbs are illuminated after the 100th person has passed through the room?

 

[MurAtt]

Will skip this one ...

Though on logically thinking its very easy ....

Since only multiples have to be found out ... 64=divisible by 2 4 8 i.e. 2x4x8=64 so all multiples of 2,4,8 will also flip the switch ...

So 1 On, 2 Off, 4 On, 8 Off, 16 On, 32 Off, 64 On ....

Answer ON





Elementary ... chuck the dirt out .... so it is in trading .. if you are able to chuck the dirt out then you will hit gold in all of your trades. :thumb:

 

[rkkarnani]

Will skip this one ...

Though on logically thinking its very easy ....

Since only multiples have to be found out ... 64=divisible by 2 4 8 i.e. 2x4x8=64 so all multiples of 2,4,8 will also flip the switch ...

So 1 On, 2 Off, 4 On, 8 Off, 16 On, 32 Off, 64 On ....

Answer ON


Elementary ... chuck the dirt out .... so it is in trading .. if you are able to chuck the dirt out then you will hit gold in all of your trades. :thumb:

SM Bhai, you made it sound very easy, but the Quiz says, No.1 switches on every bulb, no.2 changes the position of every second bulb, no.2 4 6 , and number three does this with no.3,6,9..... No. 51 onwards get to handle only a single switch.. the 50th changes the position of 50th and 100th and 51st person does it for only 51st switch.... my view is that there is more to it than merely even odd numbers....

Buti agree to your first line : Will skip this one!!!

 

[MurAtt]

Will skip this one ...

Though on logically thinking its very easy ....

Since only multiples have to be found out ... 64=divisible by 2 4 8 i.e. 2x4x8=64 so all multiples of 2,4,8 will also flip the switch ...

So 1 On, 2 Off, 4 On, 8 Off, 16 On, 32 Off, 64 On ....

Answer ON





Elementary ... chuck the dirt out .... so it is in trading .. if you are able to chuck the dirt out then you will hit gold in all of your trades. :thumb:

SM Bhai, you made it sound very easy, but the Quiz says, No.1 switches on every bulb, no.2 changes the position of every second bulb, no.2 4 6 , and number three does this with no.3,6,9..... No. 51 onwards get to handle only a single switch.. the 50th changes the position of 50th and 100th and 51st person does it for only 51st switch.... my view is that there is more to it than merely even odd numbers....

Buti agree to your first line : Will skip this one!!!

Yes so you take multiples coz sa if it has been a no 9 switch then no 9 would have been flipped by no 1, no 3 and no 9 no one else.
Same with 64

It can be flipped only by 1, 2, 4, 8, 16, 32 and 64 and no-one else.
The position of the switch remains same when others come and go as they do not fall in line of 64.

Like taking live eg no 3 will not change order of 64 as he will be switching 60, 63, 66 and so forth ... same with 5, 6, 7 BUT 8x8=64 so no 8 will flip the switch ...

Continue on and as I have serially written above and repeating again here ...
1 On, 2 Off, 4 On, 8 Off, 16 On, 32 Off, 64 On ....

QED

 

[rkkarnani]

Yes so you take multiples coz sa if it has been a no 9 switch then no 9 would have been flipped by no 1, no 3 and no 9 no one else.
Same with 64

It can be flipped only by 1, 2, 4, 8, 16, 32 and 64 and no-one else.
The position of the switch remains same when others come and go as they do not fall in line of 64.

Like taking live eg no 3 will not change order of 64 as he will be switching 60, 63, 66 and so forth ... same with 5, 6, 7 BUT 8x8=64 so no 8 will flip the switch ...

Continue on and as I have serially written above and repeating again here ...
1 On, 2 Off, 4 On, 8 Off, 16 On, 32 Off, 64 On ....

QED

Mathematics was not my strong point ever.... yes, agreed.. what about the next part... Let us skip it....

 

[smartboy1jan1982]

Question: 100 bulbs

There are 100 light bulbs lined up in a row in a long room. Each bulb has its own switch and is currently switched off. The room has an entry door and an exit door. There are 100 people lined up outside the entry door. Each bulb is numbered consecutively from 1 to 100. So is each person. Person No. 1 enters the room, switches on every bulb, and exits. Person No. 2 enters and flips the switch on every second bulb (turning off bulbs 2, 4, 6, ). Person No. 3 enters and flips the switch on every third bulb (changing the state on bulbs 3, 6, 9, ). This continues until all 100 people have passed through the room.

What is the final state of bulb No. 64? And how many of the light bulbs are illuminated after the 100th person has passed through the room?

Light Bulb 64 is on. The total number of bulbs which are on including number 64 is 10

 

[MurAtt]

Ohh .. i was so engrossed with 64 I never knew a second question exited ..

 

[smartboy1jan1982]

First think who will operate each bulb, obviously person no. 2 will do all the even numbers, and say person no. 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 & 36, 2 & 13, 6 & 6

Well in this case whilst all the factors are in pairs the number 6 is paired with it's self. Clearly the sixth person will only flick the bulb once and so the pairs don't cancel. 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.

 

[vssoma]

:clap: :clap: :thumb: :rofl:

 

[vssoma]

Question: 100 bulbs

There are 100 light bulbs lined up in a row in a long room. Each bulb has its own switch and is currently switched off. The room has an entry door and an exit door. There are 100 people lined up outside the entry door. Each bulb is numbered consecutively from 1 to 100. So is each person. Person No. 1 enters the room, switches on every bulb, and exits. Person No. 2 enters and flips the switch on every second bulb (turning off bulbs 2, 4, 6, ). Person No. 3 enters and flips the switch on every third bulb (changing the state on bulbs 3, 6, 9, ). This continues until all 100 people have passed through the room.

What is the final state of bulb No. 64? And how many of the light bulbs are illuminated after the 100th person has passed through the room?

Solution:

Light Bulb 64 is on. The total number of bulbs which are on including #64 is 10.

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 will 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 & 36, 2 & 13, 6 & 6 Well in this case whilst all the factors are in pairs the number 6 is paired with it's self. Clearly the sixth person will only flick the bulb once and so the pairs don't cancel. 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.