Author Topic: Riddle Me This...  (Read 9544 times)

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Offline Black_Magic

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Re: Riddle Me This...
« Reply #120 on: July 21, 2008, 03:36:55 AM »
I edited my post above because I was working on it before you posted - and didn't want to double post.  I solved it.

I almost hate you Glenghis, I've stayed up way past my bed time trying to solve this puzzle. I'm going to bed!


I haven't started the 2nd one yet.
« Last Edit: July 21, 2008, 03:45:00 AM by Black_Magic »
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Offline Black_Magic

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Re: Riddle Me This...
« Reply #121 on: July 21, 2008, 12:26:48 PM »
Ok.  For #2 -  A doesn't know so that means B != C and B doesn't know so that means A !=C   and that both of B's possibilities cannot be C.  I was going to say that A != B from the communitive (I think?) property - but that doesn't make sense, just because B != C and A != C doesn't mean that A != B.  However, since C = 5, it doesn't matter - A != B in this situation :P

C = A+B or C = |A-B|

Hmm.
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Offline Black_Magic

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Re: Riddle Me This...
« Reply #122 on: July 21, 2008, 03:02:23 PM »
Is anybody else trying to solve this, or am I all alone here? :(
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Offline Alex

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Re: Riddle Me This...
« Reply #123 on: July 21, 2008, 04:13:50 PM »
you're all aloooooooone!!!! booooooooooooo.
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Offline Jim S

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Re: Riddle Me This...
« Reply #124 on: July 21, 2008, 05:19:17 PM »
I'll take a shot at it....but not until tonight at the earliest. 
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Offline seanahan

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Re: Riddle Me This...
« Reply #125 on: September 15, 2008, 12:46:35 AM »
So, can we get an answer here?
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Offline Glenghis Khan

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Re: Riddle Me This...
« Reply #126 on: September 20, 2008, 07:36:04 PM »
So, can we get an answer here?

Oh right, sure :). The answer is:

A=2
B=3
C=5

Explanation:

A sees 3 and 5 so can only deduce that his number is either 8 (5+3) or 2 (5-3)

B sees 2 and 5 so can only deduce that his number is either 7 (5+2) or 3 (5-2), and he can't learn anything new from the fact that A does not know his number since A would not be able to figure out his number regardless of what B's number was.

C knows that he is either 1 or 5.  If C was 1 then he knows that A would be able to work out that he was 2 or 4 and B would be able to work out that he was 1 or 3.  B would know that he couldn't be 1 however since this would mean A would see two 1's and be able to deduce tht he was 2 immediately.  Therefore B could deduce that he was 3 if C was 1.

C in reasoning through all of this, knows that he must be 5 and not 1.

This was hard.  I don't blame anyone for not trying, perhaps I should have given the numbers and had people work out why they were what they were.


Offline Black_Magic

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Re: Riddle Me This...
« Reply #127 on: September 21, 2008, 12:11:26 AM »
You can't have the numbers already and then explain why - that's a guess and check deal.

I call bullshit.
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Offline Glenghis Khan

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Re: Riddle Me This...
« Reply #128 on: September 21, 2008, 12:19:11 AM »
You can't have the numbers already and then explain why - that's a guess and check deal.

I call bullshit.
I dont know a method to find the numbers - there probably isn't one that doesnt involve some guesswork.

I cheated for the second part and looked at the answer and then worked out why is was what it was.

Offline Black_Magic

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Re: Riddle Me This...
« Reply #129 on: September 21, 2008, 12:30:56 AM »
There has to be a way.

You can't leave us hanging like that.









Or Black_Magic will be unhappy :(
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Offline Jim S

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Re: Riddle Me This...
« Reply #130 on: September 21, 2008, 01:55:29 AM »
Oh, I never came back to this.   

And...you don't want Black Magic mad at you.    It is bad.  ;)
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Offline Glenghis Khan

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Re: Riddle Me This...
« Reply #131 on: September 21, 2008, 02:43:40 AM »
Oh, I never came back to this.   

And...you don't want Black Magic mad at you.    It is bad.  ;)

Oh no!

Well B!=C and A!=C is clear.

Also B cannot be any of A's options or A would be able to eliminate one and know his number, so B!=B+C (true for all B anyway) and B!=|B-C| so B is not 2C or C/2 (which it cant be anyway).

Similarly since B does not know his number, A!=2C.

We have three cases:
1)A+B=C  (A and B are two positive integers that sum to 5)
2)A+C=B  tells us B!=3C
3)B+C=A  tells us A!=3C

By iiteration A and B cannot be multiples of C. 

Case 2) and 3) imply |A-B|=5

Then its a simple matter of showing that |A-B|!=5 reducing the problem down to two simple options A and B are 1 and 4 or 2 and 3.

>_> 
<_<



« Last Edit: September 21, 2008, 02:57:45 AM by Glenghis Khan »

Offline Glenghis Khan

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Re: Riddle Me This...
« Reply #132 on: September 21, 2008, 02:49:44 AM »
A!=B is true also because if they were both larger than C then C would have to be zero and C does not divide evenly into 2 for them to be both smaller.

Offline Black_Magic

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Re: Riddle Me This...
« Reply #133 on: September 21, 2008, 02:24:26 PM »
whoooooaaa boy.


Slow down there... I haven't looked at these problems in months.


Explain it again.


This time in English.
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Offline Glenghis Khan

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Re: Riddle Me This...
« Reply #134 on: September 21, 2008, 06:37:59 PM »
What I wrote could easily be garbage, it came to me in a late night epiphany, I'll possibly revise it later.

 

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