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Uhhh oxygen?

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"Hit hard, hit first, hit often."

--Adm. William "Bull" Halsey, USN

Tommorow

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If you've been mod-o-fied,
It's an illusion, and you're in-between.
Don't you be tarot-fied,
It's just alot of nothing, so what can it mean?
~Frank Zappa

Sins exist only for people who are on the Way or approaching the Way

Wow, that was fast. *sigh*

Mogmaster, you're up.

Alright, no one's posted a new one in a couple of days, so I'm taking it.

A school has 1000 lockers, numbered 1-1000. On the first day of school, all the kids line up outside. The first kid runs through the school and opens every locker. The second kid runs through and closes every even-numbered locker. The third kid comes through and changes the state of every 3rd locker (if it's open, he closes it, if it's closed, he opens it.) The fourth child changes every 4th locker, and so on until 1000. At the end of this process, how many lockers are open? And which ones?

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Hip-Hop QOTW:

"Yeah, where I'ma start it at, look I'ma part of that
Downtown Philly where it's realer than a heart attack
It wasn't really that ill until the start of crack
Now it's a body caught every night on the Almanac"

"Game Theory"
The Roots

Is it: One locker is open and its the first one?

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<Nealio> Seymour sounds like evil Winnie the Pooh

Does "The fourth child changes every 4th locker, and so on until 1000." mean that there are 1000 children, and the 1000th child will change every 1000th? Because if it's just four kids, the answer is crazy. I wrote a java program that outputted the answer...and it's not really postable.

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Hey, put the cellphone down for a while
In the night there is something wild
Can you hear it breathing?
And hey, put the laptop down for a while
In the night there is something wild
I feel it, it's leaving me

Surely he means the first kid opens them all, the second kid closes every second one, third kid swaps every third one, fourth kid swaps every fourth one, and so on and so on.

At least I think that's what he means.

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Little miss Muffet, sat on her tuffet,
Eating her curds and...
WHEEEEEEEEEEEEEEEEEEEEEEEEEEE!

If that's the case, then the ones that are open are 1, 4, 9, 16, and every perfect square up to 31^2, which is 961.

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Hey, put the cellphone down for a while
In the night there is something wild
Can you hear it breathing?
And hey, put the laptop down for a while
In the night there is something wild
I feel it, it's leaving me

Got an explanation? Not one that goes, "I simulated it using a Java program."

Edit: Ooh, it makes sense! Brilliant!

This post has been edited by Goobers on 24th March 2004 20:07

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Little miss Muffet, sat on her tuffet,
Eating her curds and...
WHEEEEEEEEEEEEEEEEEEEEEEEEEEE!

Unfortunately, I don't have an explanation. I simulated it using a Java program.

Code

public class Gears
{
public static void main(String[] args)
{
 boolean[] lockers = new boolean[1001];
 // Makes an array from 0 to 1000.  0 will be ignored in my code
 for (int i = 1; i < lockers.length; i++)
 {
  // true = open
  // false = closed
  // this code sets all lockers open, like the first kid
  lockers[i] = true;
 }
 for (int h = 2; h < lockers.length; h++)
 {
  // h is the kids - starts from two, since the first kid already opened them all
  // i is the lockers - starts from 1 and goes to 1000
  // this double loop sends every kid past every locker
  for (int i = 1; i < lockers.length; i++)
  {
   if (i % h == 0)
   {
    if (lockers[i] == false)
    {
     lockers[i] = true;
    }
    else
    {
     lockers[i] = false;
    }
   }
  }
 }
 for (int i = 1; i < lockers.length; i++)
 {
  System.out.print(i);
  System.out.println(": " + lockers[i]);
 }
}
}

This post has been edited by Neal on 24th March 2004 20:13

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Hey, put the cellphone down for a while
In the night there is something wild
Can you hear it breathing?
And hey, put the laptop down for a while
In the night there is something wild
I feel it, it's leaving me

I think I understand. Think of it this way:

Every number has an even number of factors. This should be obvious, because it takes TWO numbers to multiple to get another one. So for the number 6, in order for 2 to be a factor, 3 must also be a factor. Now, an even number of locker door swapping results in closed. (open, closed, open, closed, open, closed). Take the number 7. In order for 1 to be a factor, 7 has to be a factor. Take the number 12, 1 x 12, 3x4, 2x6.

BUT, what if those two factors were actually the same factor? This is why the perfect squares are open. 49 has 1x49, 7x7. But 7 twice is the same thing. So the locker door swaps an ODD number of times.

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Little miss Muffet, sat on her tuffet,
Eating her curds and...
WHEEEEEEEEEEEEEEEEEEEEEEEEEEE!

Great explanation, Goobers. I think Neal got it first, though.

--------------------

Hip-Hop QOTW:

"Yeah, where I'ma start it at, look I'ma part of that
Downtown Philly where it's realer than a heart attack
It wasn't really that ill until the start of crack
Now it's a body caught every night on the Almanac"

"Game Theory"
The Roots

I'll hand it over to Goobers, though, since he provided the awesome explanation for it, and I used a program to figure it out. That, and I'm plumb out of riddles at the moment.

--------------------

Hey, put the cellphone down for a while
In the night there is something wild
Can you hear it breathing?
And hey, put the laptop down for a while
In the night there is something wild
I feel it, it's leaving me

Hmm...

Let's play a game. I'm going to put 100 quarters on the table. Twenty of them are tails, the rest are heads. Now, if you split these quarters into two groups such that there is an even number of tails in both piles, you win all the quarters!

The catch? You're blindfolded and wearing gloves so that you can't feel whether the quarters are tails or heads.

How do you do win my 25 bucks?

EDIT: You do not see my quarters beforehand, nor are they in separate piles yet. It's just one big mass of quarters, and you come into this problem completely blind.

This post has been edited by Goobers on 27th March 2004 20:52

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Little miss Muffet, sat on her tuffet,
Eating her curds and...
WHEEEEEEEEEEEEEEEEEEEEEEEEEEE!

I think i get it, and I have two solutions.
1: you seperate them into piles beforehand so it is simply remembering wich ones are in what pile and then it is easy to sort.

OR

2: you just put them in piles of 50 each...each quarter has both a head and a tail so as long as the pile are equal they will have an even number of each.

And besides nobody wants $25 in quarters!

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There is a time and a place for everything, and I am niether

I got it!

You pick up the quarters, and you lick the bottom, and get an idea of the shape on the bottom, so you know which side was the top!

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"And that, my liege, is how we know the Earth to be banana-shaped” -Sir Bedevere the Wise

Nice try, Fadien. I want to give it to you just for creativity by suggesting the licking of the quarters.

But no.

Keep going, guys.

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Little miss Muffet, sat on her tuffet,
Eating her curds and...
WHEEEEEEEEEEEEEEEEEEEEEEEEEEE!

this is pretty simple. seperate 50 coins off to form one pile, and then turn all 50 over.

In response to Goober's Edit:
Ok you seperate them into piles of 50 and then that is all. Assuming they are all normal quarters each pile will have 50 heads and 50 tails because each quarter has one of each.




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There is a time and a place for everything, and I am niether

Err, let me re-tell the riddle. There are 100 quarters on the table. 20 are tails, and I want you to separate them into two groups. The number of tails in one group has to equal to the number of tails in the other group. You cannot see or feel (or lick ) the quarters.

No one has gotten the answer yet.

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Little miss Muffet, sat on her tuffet,
Eating her curds and...
WHEEEEEEEEEEEEEEEEEEEEEEEEEEE!

So are you saying that only 20 of them HAVE tails? I think you need to clarify as to how the coins are laid out at the begining.


POST NUMBER 100

This post has been edited by The Gambler on 2nd May 2004 21:45

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There is a time and a place for everything, and I am niether

Argh! There's no trick in my wording. These are 100 normal quarters, 20 of them are tails side up, 80 are heads side up. Now, I want you to split them into 2 piles such that the number of tails side up quarters are equal in both piles.

If a majority of the people start asking for the solution, I'll post it.

This post has been edited by Goobers on 28th March 2004 21:12

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Little miss Muffet, sat on her tuffet,
Eating her curds and...
WHEEEEEEEEEEEEEEEEEEEEEEEEEEE!

just seperatwe them both into piles of 50
and be content without $25 dollars added to your wallet



This post has been edited by Cloud_Strife510 on 29th March 2004 02:11

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"Have you ever seen a baby do that before?"

I wish I had 25 dollars in my wallet... but I don't, and that's why I make sure I can win this game!

After I give you the solution, you can too!

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Little miss Muffet, sat on her tuffet,
Eating her curds and...
WHEEEEEEEEEEEEEEEEEEEEEEEEEEE!

whoops, i typed '50' when i meant '20.' :-\ i guess i had the gamber's solution in my head. so, take off 20 coins to form one pile, and flip all 20.

Gozaru has it.

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Little miss Muffet, sat on her tuffet,
Eating her curds and...
WHEEEEEEEEEEEEEEEEEEEEEEEEEEE!

let's do another math riddle.

raynjerz decides to throw a CoN party and invites all 50 (bear w/ plz) active forum/chat members. so when the party day comes around, all 50 people come, and all 50 bring their girlfriends/boyfriends. so everyone has a good time etc etc. before everyone leaves, raynjerz asks everyone and their spouse (including angie) to put in a box a slip of paper written down how many people's hands they shook. when he read throgh the papers, he noticed that *every number* from 0 to 100 was listed (note: with 101 submisions, this obviously means that no number was repeated). ifno one shook their partner's hand, how many people's hands did raynjerz shake?

-edit- this 101 is non-inclusive of raynjerz. i thought it was obvious, but the 101 is for the 50 people, their partners, and angie. raynjerz did not submit his information (he doesn't remember, but he can find out from the results).

This post has been edited by gozaru~ on 31st March 2004 06:09

It's probably too obvious to be right, but 100? There are 101 people there (50 members, 50 b/gfs and raynjerz), and if no one shook their parter's hand (or, obviously, thier own hand) the max anyone else could have would be 99. So...raynjers shook 100.

I've looked at this every way I can, and I can't see how it is possible to solve.


You see, there is 102 people at the party. Raynjer, Angie, and 50 people and their partner.

Now, if every person didn't shake their partners hand, and can't shake theirs, that leaves 100 people for them to shake hands with.


So... if one person shook 100 other hands, then their partner would have shook 0. So Raynjer shook atleast that person's hand, and that leaves 3 hands he couldn't have shook (His, angie, and the person that shook 0.)


Now, the person who shook 1 hand, had to have shaken the guy who shook 100 hands, so that totals 1 raynjer shook, and 4 he hasn't, and the person that shook 99 didn't shake 1 or 0, meaning they DID shake raynjer's hand... so.. 2 he did, 4 he didn't.. and the person who shook 2 shook 100 and AH HA!! I SEE HOW IT WORKS! *does some calculations*


*10 min later*

Hmmmm...

*10 more minutes*

Raynjer shook 50 hands, 51 (52 including himself) he didn't shake. He shook hands with those who shook 51 through 100 hands, because 50 shook hands with...

Well.. 51 shook hands with 50, 52-100, and Ranjer... 49 shook hands with 52-100, and 52 shook hands with 51, 49, 53-100, and Raynjer... and....

Brain hurts.... can't explain....


So my final answer is he shook hands with 50 people. The person who shook hands with 51, up to the person who shook hands with 100....

Yea...

This post has been edited by Fadien on 31st March 2004 07:29

--------------------

"And that, my liege, is how we know the Earth to be banana-shaped” -Sir Bedevere the Wise

correct. the answer is 50. good job on solving it.