Puzzler...

[Lucretius]

Heard this one a while ago, thought I'd share it with you fellas...

Imagine you have a very long hall with 20,000 lightbulbs in line on it's ceiling. Now imagine one person goes through and turns each one on. Then a second guy goes through and pulls the swith of every other bulb (2,4,6,8...). Then a thrid pulls the switch of every third bulb(3,6,9,12...). This process repeats untill the 20,000th person does every 20,000th bulb (ie, just the last one).

Questions:
1) Will this bulb be on or off?
a)4
b)12
c)26
d)16,641

2) How do you know?

3) Which bulbs will be on?

[sam]

Is luc back? have I missed an announcement?

[gannon]

Don't worry, it's only a ghost

[Turbo Tax 1.0]

d) proccess of elimination the one guy turns off all of the even bulbs thus leaving the one odd one on
unless its a trick question and none of the choices are possible

[soundwave]

But if you add up the digits of 16641, it equals 18, meaning its divisible by 3, so itd be turned off.

[daguuy]

i think i got it:
1)
a)off
b)off
c)off
d)off

2) the 1st guy got rid of all the multiples of 2 (all but d) and d is a multiple of 3 (picked by the 3rd guy)

3) only the 1st one would be on because people picked 2+. the first guy (the only one who would get the 1st one) put them in, not turn them off

[soundwave]

a) on (turned back on by #4.
b) on (turned on by 4, off by 6, on by 12)
c) off (off by 2, on by 13, off by 26)
d) a little too big.

[Turbo Tax 1.0]

But if you add up the digits of 16641, it equals 18, meaning its divisible by 3, so itd be turned off.

dang i thought that one had to be on o well

i cant believe i forgot about those divisible rules looks like im just forgetting more and more as i go

[nos_slived]

a) on (turned back on by #4.
b) on (turned on by 4, off by 6, on by 12)
c) off (off by 2, on by 13, off by 26)
d) a little too big.

You forgot the second and third people in (b). On by 1, off 2, on 3, off 4, on 6, and off by 12.

[Lucretius]

Ready for an answer?

1a) On
1b) Off
1c) Off
1d) On

2) You're looking at factors. Take 26. How many times will anyone touch it?
Guy 1, 2, 13, and 26. No one else. So that means it's turned on, off, on, off. Try it with the rest.

3) This will help with the big ones. Only squares will be on, for only they have an odd number of people touching them. Ie: 9- 1, 3, 9. (3 factors)

But say, 32-1, 2, 4, 8, 16, 32. (6 factors)

So how do you know if a bulb is on? Take the square root of it's number. If you get a whole number, it's on. If you don't, it isn't.

Fun, huh?

[nos_slived]

I thought it might be something like that, but then I started thinking in terms of prime numbers, which obviously wasn't right, and then I forgot about my earlier thought.

Can you tell it's 2am? Usually if my posts make slightly less sense(or sometimes more) than usual, it's probably really late. Or if I start humping a lot of extra legs...

[Lucretius]

I thought my shin was a little dry...

[rawls]

2) You're looking at factors. Take 26. How many times will anyone touch it?
Guy 1, 2, 13, and 26. No one else. So that means it's turned on, off, on, off. Try it with the rest.

3) This will help with the big ones. Only squares will be on, for only they have an odd number of people touching them. Ie: 9- 1, 3, 9. (3 factors)

But say, 32-1, 2, 4, 8, 16, 32. (6 factors)

So how do you know if a bulb is on? Take the square root of it's number. If you get a whole number, it's on. If you don't, it isn't.

You have the right idea, but I feel your explanation is somewhat lacking. Essentially every number X is going to have N factors, where N is divisible by two. The reason N is divisible by two is that every factor N1 will have a "pair" factor, N2, such that N1*N2=X. Let's look at 32. The (N1, N2 ) pairs are:

(1,32 ), (2, 16 ), (4, 8 ).

For a prime number, there are only two (e.g. 11: (1, 11 )). The trick here is that squares have one factor pair with the same number, so for 36:

(1, 36 ), (2, 18 ), (3, 12 ), (4, 9 ), (6, 6 ).

While there are 5 pairs here, what matters for this problem is the number of different numbers (because each number corresponds to one guy going through and flipping switches, and though (6, 6 ) contains two numbers, guy #6 can only flip switch #36 once). Thus, as The Spoiler said above, there is only one way for you to get an odd number of guys hitting a particular switch: if that particular switch's number is a perfect square. However, the reason for this is not as trivial as he made it sound.