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icon3.gif  Puzzle thread May 3, 2011 Tue, 03 May 2011 17:11 Go to next message
Ron is currently offline Ron

 
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It's been 6 years since I've done a puzzle thread.

Here goes. 24 hour time limit, PM me your answers. If you answer 5 questions correctly, you get 1 chance to be the Puzzle Master. If you get all 6 questions correct, you get 2 chances.

The thread will be locked until then.

1. Four chess players, Fred, Stan, Dan, and Steve competed in a tournament. At the end of the tournament, they said the following:

Fred: "I did not win 1st or 4th place."
Steve:"I got 4th place."
Stan: "I did not get 4th place."
Dan: "I won 1st place."

One, and only one of the chess players is not telling the truth. Who won the tournament? How can you tell?

2. The cost of 175 pens is greater than that of 125 pencils, but less than that of 126 pencils. The cost of one pen is a whole number of cents, as is the cost of one pencil. If you wanted to buy 1 pencil and 3 pens, would 100 cents be enough? (Show your work.)

3. What is the highest number of pawns that can be placed on a chess board so that no four pawns are in a line? Lines to avoid are verticals, horizontals, and all types of diagonal. No pawn may be isolated, all pawns must be a neighbor with at least one other pawn either vertically or horizontally. (use the CODE tag and Courier font to diagram your answer)

4. Which number is the odd one out and why?

3628, 2426, 4146, 1448

5. The diagram below features 4 houses, labeled C, X, Y, and Z. One of the houses, C, is located at the center of a circle, and the other 3 houses, X, Y, and Z are somewhere on the circumference. There are paths connecting the various houses as shown. A member of family Y decides to take a walk starting from his house going clockwise around the circular path. Meanwhile, another member of family Y walks along the straight path, visiting houses X, C, and Z in order, but not stopping, before returning home by the direct ZY path.

Assuming that the two members of family Y walk at the exact same rate, who arrives home first?
http://starsautohost.org/puzzle_cl.jpg
...




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Re: Puzzle thread May 3, 2011 Wed, 04 May 2011 21:46 Go to previous messageGo to next message
Ron is currently offline Ron

 
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Only 2 people replied. I'm sorry that I was not able to go through them today, but will go through them tomorrow.


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Re: Puzzle thread May 3, 2011 Thu, 05 May 2011 10:16 Go to previous messageGo to next message
Ron is currently offline Ron

 
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Ok, here we go. Only 2 users submitted answers for this Puzzle Thread, Gible and Azrael. Technically, they only got 4 correct answers each, but I gave them each #4 as each of them submitted multiple possible answers and reasoning to support each answer for #4.

Here are the answers.

1. The winner was Stan, and Dan was lying. With only Dan lying, Fred, Dan, and Steve couldn't have won. All other possibilities don't work in one way or the other.

2. Let the cost of one pen be D cents and the cost of one pencil be C cents. Therefore, 125C < 175D < 126C. This in turn means that 5C < 7D and 25D < 18C, so that 5C + 1 <= 7D and 25D <= 18C - 1. If we multiply these two inequities by 25 and 7 respectively, we obtain 125C + 25 <= 175D <= 126C - 7. This implies that 125C + 32 <= 126C and C >= 32. Now 7D >= 5C + 1 >= 161, or D >= 23. So C + 3D >= 32 + 3(23) = 101 cents, which means that 100 cents is not enough to buy one pencil and 3 pens.

3. 15 pawns can be placed on a chess board so that no more than 3 are in a line (horizontal, vertical, or diagonal). All pawns must have at least one neighbor next to it either horizontally or vertically.
[FONT=Courier]
.......
ppp....
..ppp..
....ppp
.p.pp..
.ppp...
[/FONT]


4. This question was in the Mensa section. Since both users gave multiple alternate answers and reasons for each of the 4 numbers I gave both of them this one free.

The answer I was looking for was:
4146: In all the others, AxC=B and B+C=D.
For example:
3628
ABCD

5. Note that the distances XY and YZ must each be less than the diameter of the circle. Also, segments XC and CZ must sum to precisely the diameter of the circle, because each one of these two segments is a radius.

Therefore, the sum of all four of these segments must be less than three times the diameter of the circle. But the circumference equals pi times the diameter, and pi is greater than 3, so the family member who walks along the straight paths arrives home first.

6.
...



[Updated on: Thu, 05 May 2011 10:18]




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Re: Puzzle thread May 3, 2011 Thu, 05 May 2011 10:44 Go to previous messageGo to next message
Azrael is currently offline Azrael

 
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I dont understand your answer for Question 3.
A normal Chess Board has 8*8=64 fields. Yours only have 6*7=42
On a 8*8 Board you can put at least 4 Blocks with nine Pawns in each block. With a litle trying you can put ther some more on it.

And those number things are very unclear.

The first Thing i learned in my Math Studys is that if you have something like 1,2,3,4,... then evrything can happen. If you dont have the definition of a set, you dont know wich Numbers are in it.

But i liked the Questions. Does anyone know how to prove the thing with the Pawns?


[Updated on: Thu, 05 May 2011 10:45]

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Re: Puzzle thread May 3, 2011 Thu, 05 May 2011 11:19 Go to previous messageGo to next message
slimdrag00n is currently offline slimdrag00n

 
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I admit I gave up after thinking of #1. What is the logic to finding out who is telling the truth or not? Good guessing?

The rest of the questions you have to know math like the back of your hand so its no wonder only two people took on this challenge.




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Re: Puzzle thread May 3, 2011 Thu, 05 May 2011 11:41 Go to previous messageGo to next message
Azrael is currently offline Azrael

 
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slimdrag00n wrote on Thu, 05 May 2011 17:19

I admit I gave up after thinking of #1. What is the logic to finding out who is telling the truth or not? Good guessing?

The rest of the questions you have to know math like the back of your hand so its no wonder only two people took on this challenge.




Its simple. Think only about what would it if ### lies? Then you will come to the conclusion that at least one other person have to lie too. So only on of them can lie without forcing someone other to lie.

You know then who of them is the one that lies. Then there is only one Person that can be the Winner.

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Re: Puzzle thread May 3, 2011 Thu, 05 May 2011 11:52 Go to previous messageGo to next message
m.a@stars is currently offline m.a@stars

 
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slimdrag00n wrote on Thu, 05 May 2011 17:19

I admit I gave up after thinking of #1. What is the logic to finding out who is telling the truth or not? Good guessing?

Not so different from guessing PRTs in a game, after all. Sherlock Deal



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Re: Puzzle thread May 3, 2011 Thu, 05 May 2011 12:01 Go to previous messageGo to next message
slimdrag00n is currently offline slimdrag00n

 
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Fred: "I did not win 1st or 4th place."
Steve:"I got 4th place."
Stan: "I did not get 4th place."
Dan: "I won 1st place."

But how do you know stan lied? He says he did not get 4th place. He could have got 3rd place no?

Dan says he got 1st place but noone else said they got first place so this could be true.

Steve said he got 4th place but he could have got 2nd or 3rd?

Fred could have won 1st or 4t?

I am just not understanding the exact logic of how to tell who is really being truthful. I mean the answer could be that they are all liars?



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Re: Puzzle thread May 3, 2011 Thu, 05 May 2011 12:32 Go to previous messageGo to next message
Azrael is currently offline Azrael

 
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slimdrag00n wrote on Thu, 05 May 2011 18:01

Fred: "I did not win 1st or 4th place."
Steve:"I got 4th place."
Stan: "I did not get 4th place."
Dan: "I won 1st place."

But how do you know stan lied? He says he did not get 4th place. He could have got 3rd place no?

Dan says he got 1st place but noone else said they got first place so this could be true.

Steve said he got 4th place but he could have got 2nd or 3rd?

Fred could have won 1st or 4t?

I am just not understanding the exact logic of how to tell who is really being truthful. I mean the answer could be that they are all liars?


If Stan would lie than also Steve have to lie. So Stan cant lie.
If Fred would lie than also Steve or Dan have to lie. So Fred cant lie.
If Steve would lie than also someone of the others have to lie (for first place also Dan have to lie, for second and third Fred or Stan have to lie)
So Dan is the only on that can lie, if only one of them lies.

So now you know that Dan lies, and the only one that can have won is Stan

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Re: Puzzle thread May 3, 2011 Thu, 05 May 2011 12:39 Go to previous messageGo to next message
slimdrag00n is currently offline slimdrag00n

 
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Thanks for breaking it down. Now I understand the elimination process.


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Re: Puzzle thread May 3, 2011 Thu, 05 May 2011 13:06 Go to previous messageGo to next message
nmid

 
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darn.. dunno how I missed this.

When is the next one coming out? I would like to try my hand* too.

* - Brain Wink


[Updated on: Thu, 05 May 2011 13:06]




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Re: Puzzle thread May 3, 2011 Thu, 05 May 2011 15:05 Go to previous messageGo to next message
dennis is currently offline dennis

 
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Yeah Ron..whens the next one....donjon sent me an email telling me to get a life and pointed me to this thread...lol.....

Its been ages....greetings fellow beings.



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Re: Puzzle thread May 3, 2011 Thu, 05 May 2011 16:09 Go to previous messageGo to next message
Ron is currently offline Ron

 
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nmid wrote on Thu, 05 May 2011 13:06


When is the next one coming out? I would like to try my hand* too.

* - Brain Wink

We used to call these Weekly Puzzle Threads when I ran them starting late Oct. 2003 through Jan. 2004, and one in 2005.

They're a fair amount of work to run, so doubt I could run one every week, but I'll try to do another one next week. Part of the work is trying to ensure that the puzzles cannot be easily Googled.

Do a search for Puzzle Thread to look back over the old ones and try your hand on them also.



[Updated on: Thu, 05 May 2011 16:10]




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Re: Puzzle thread May 3, 2011 Tue, 07 June 2011 15:35 Go to previous message
dennis is currently offline dennis

 
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Try This or These



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