Random Riddles

#4 Yes you can.
1. Weigh eight balls. Four on each side of the scale. If they weigh the same go to step 1a, if they don't go to step 1b

1a. Now you know that none of those eight balls is the odd one so put them away. Weigh the last four balls against each other. Two on each side of the scale. One of the sides will tip down. That side contains the odd ball. Take those two balls and weigh them against each other and you know wich one's the odd ball.

1b. Take the four balls on the side that tipped down and weigh them against each other. Two on each side of the scale. One of the sides will tip down. That side contains the odd ball. Take those two balls and weigh them against each other and you know wich one's the odd ball.

Don't know if it's correct. It makes sense to me, but then again I'm so damn tired.
 
Prodigal Son, you'd be right if it wasn't for the fact that we don't know if the odd ball is heavier or lighter.
 
[!--QuoteBegin-LooseCannon+Dec 12 2004, 08:52 PM--][div class=\'quotetop\']QUOTE(LooseCannon @ Dec 12 2004, 08:52 PM)[/div][div class=\'quotemain\'][!--QuoteEBegin--]#2.  They drove backwards.
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Nice! [!--emo&B)--][img src=\'style_emoticons/[#EMO_DIR#]/cool.gif\' border=\'0\' style=\'vertical-align:middle\' alt=\'cool.gif\' /][!--endemo--] One down...
 
[!--QuoteBegin-Prodigal_Son+Dec 12 2004, 07:52 PM--][div class=\'quotetop\']QUOTE(Prodigal_Son @ Dec 12 2004, 07:52 PM)[/div][div class=\'quotemain\'][!--QuoteEBegin--]#4 Yes you can.
1. Weigh eight balls. Four on each side of the scale. If they weigh the same go to step 1a, if they don't go to step 1b
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And if they don't weigh the same, which is more likely in the first place?

I studied problems similar to this in a math class this past term, and I think #4 is impossible. However, I'll keep searching for a solution anyway.

EDIT: I found two ways to do #4 in 4 weighings, but I can't get it down to 3 yet...
 
[!--QuoteBegin-LooseCannon+Dec 13 2004, 12:27 PM--][div class=\'quotetop\']QUOTE(LooseCannon @ Dec 13 2004, 12:27 PM)[/div][div class=\'quotemain\'][!--QuoteEBegin--]#3: Seventh floors on hotels never have balconies.
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I thought of something similar lol. I was thinking Seventh floor ROOMS (not the entire floor) didn't have balconies... she wasn't in her room.
 
[!--QuoteBegin-LooseCannon+Dec 12 2004, 08:54 PM--][div class=\'quotetop\']QUOTE(LooseCannon @ Dec 12 2004, 08:54 PM)[/div][div class=\'quotemain\'][!--QuoteEBegin--]Prodigal Son, you'd be right if it wasn't for the fact that we don't know if the odd ball is heavier or lighter.
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That's what makes it an evil problem.
[!--QuoteBegin-Lib+Dec 13 2004, 07:44 AM--][div class=\'quotetop\']QUOTE(Lib @ Dec 13 2004, 07:44 AM)[/div][div class=\'quotemain\'][!--QuoteEBegin--]#3 : Mr jones is the murderer ? At least he's behind the murder ?
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Yes, he was the murderer.

@ Cannon, there was indeed a balcony on that seventh floor.
 
Wait, how could Mr. Jones be the murderer if he had never met the victim, as stated in the riddle? "Met" to me meant "encountered", not necessarily "spoken to" or anything like that. It seems that a murderer would have to encounter his victim if he's going to throw the victim off the balcony.

Yes, I know there are ways around this problem e.g. Jones poisoned the victim's food without meeting the victim and someone else threw the body off the balcony... but it still seems like a badly constructed riddle to me if that's the answer. Which is not the fault of SIASL of course, as I trust he found the riddle in that form somewhere.
 
Oh shit. [!--emo&:eek:--][img src=\'style_emoticons/[#EMO_DIR#]/ohmy.gif\' border=\'0\' style=\'vertical-align:middle\' alt=\'ohmy.gif\' /][!--endemo--] For some reason I thought Lib asked if Mr. Rigby-Brown was the murderer (which he was). Jones WAS NOT the murderer. Heh, sorry about that. [!--emo&:D--][img src=\'style_emoticons/[#EMO_DIR#]/biggrin.gif\' border=\'0\' style=\'vertical-align:middle\' alt=\'biggrin.gif\' /][!--endemo--]
 
Alright. How about this:

Mrs. Rigby-Brown was a midget, therefore she was unable to either fall over the railing, or ride on the roller-coasters.
 
Sorry, but no. [!--emo&:p--][img src=\'style_emoticons/[#EMO_DIR#]/tongue.gif\' border=\'0\' style=\'vertical-align:middle\' alt=\'tongue.gif\' /][!--endemo--]

I'm surprised no one has gotten 1 yet. I'd probably reveal too much by giving another hint.

2 is indeed hard. However, here is a hint:
Q: Did Jones know of the Rigby-Browns prior to reading the article?
A: Yes.

I'd rather see someone solve 4 without any hints.
 
[!--QuoteBegin-LooseCannon+Dec 13 2004, 10:28 PM--][div class=\'quotetop\']QUOTE(LooseCannon @ Dec 13 2004, 10:28 PM)[/div][div class=\'quotemain\'][!--QuoteEBegin--]Mrs. Rigby-Brown was a midget, therefore she was unable to either fall over the railing, or ride on the roller-coasters.
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Well, this was wrong but it's good to think outside the box like that. The seemingly illogical riddles I submitted can only be solved by abandoning your conventional style of thinking. Your logic.
 
I haven't been able to solve #4 yet, but I've gotten close. I can find the odd ball (and whether it is heavy or light) out of nine balls in 3 weighings. Here's how you do that:

1. Weigh 6 random balls (3 vs. 3). If they are equal, go to step 5. If not, label them such that ABC > DEF.

2. You now know that the other six balls (GHIJKL) are normal, so weigh ABCDEF vs. GHIJKL. If the ABCDEF side is heavier, go to step 3. If the ABCDEF side is lighter, go to step 4.

3. One of ABC is heavy, so weigh A vs. B. If they are equal, C is heavy; if not, the heavier of A vs. B is the heavy ball. Done in 3 tries.

4. One of DEF is light, so weigh D vs. E. If they are equal, F is light; if not, the lighter of D vs. E is the light ball. Done in 3 tries.

5. If you're on this step, then you know ABCDEF are normal. Weigh ABC vs. GHI. If they are the same, go to step 8. If ABC > GHI, go to step 6. If ABC < GHI, go to step 7.

6. One of GHI is light, so weigh G vs. H. If they are equal, then I is light; if not, the lighter of G vs. H is the light ball. Done in 3 tries.

7. One of GHI is heavy, so weigh G vs. H. If they are equal, then I is heavy; if not, the heavier of G vs. H is the heavy ball. Done in 3 tries.

8. If you're on this step, you know one of JKL is the odd ball, but you don't know if it's heavy or light. If you did know that, you could solve this puzzle with one more weighing (J vs. K).

So if you follow this method, then the riddle boils down to finding the odd ball out of 3 choices in one more weighing (remember that you have nine known normal balls to weigh against). I see no way to do this; two weighings would be required. But I have presented this method in hopes that someone may be able to improve on it to solve the riddle.
 
I gave up, Googled, and found the answer to #4 on the web. I won't post it here, but I will give the following hint: my guess in the previous post was WAY OFF. Anyone trying to improve on my previous idea is barking up the wrong tree.
 
Okay, I'll give another hint for both #1 and #3.

For 1:
This question can be answered by thinking about this: What makes god different than us? That clue alone should make the answer easy to get.

For 3:
Q: Had Jones ever provided a service for the Rigby-Browns but without meeting them?
A: Yes.

That should also be a helpful hint.
 
No, for the reason that he would have needed to meet her. Clue: Jones spoke on the phone with Mr. Rigby-Brown. I hope I'm not giving too much away with these hints and clues... [!--emo&:unsure:--][img src=\'style_emoticons/[#EMO_DIR#]/unsure.gif\' border=\'0\' style=\'vertical-align:middle\' alt=\'unsure.gif\' /][!--endemo--]
 
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