Logic question

[DarthMaximus]

I have some guesses but I can’t really pin point it yet.  Long day.

Alex = A
Bob = B
Chuck = C

When Alex first looks he has two options for his number:  B+C or B-C

He has know way of knowing which is correct.

When Bob looks he also has two options, A+C or A-C, but not enough info to determine which one.

Now Chuck looks and also has two options, A+B or A-B, but can’t decide, however, this does give Alex the required information he needs to eliminate either B+C or B-C, thus he knows he has 50.

[Avin]

Keep it up, you’re on the right track, DM. Now you just have to figure out how Alex eliminates one of his two options (B+C or B-C) on his second “turn”.

I suggest you start by considering how someone would eliminate one of their two options without hearing anything from the other two people (which Yanny seems to have gotten), then how B might have eliminated one of his two options after hearing A’s first turn, then how C might have eliminated one of his two options after hearing both A and B, and finally solving the problem itself.

[Guest]

I think that the basic problem here is this. Alex knew immediately what numbers were on the other idiot’s hats. He learned nothing after that except……I don’t know what number is on my hat, from the other 2 goofballs. That wasn’t learning anything, because Alex already knew what was on their stupid hats. problem BOGUS

[Guest]

And another thing… How in the world could Alex deduce that his number was 50 when there was no more input from the other 2 idiots than " I don’t know what my number is"?? BOGUS

[Avin]

Sorry Maddogg, as you can see I indicated that DM was clearly on the right track. If you’re interested in trying to solve the problem, you might want to consider what he was saying and try to think about it more rather than just dismissing the problem as bogus. This problem is in some ways similar to the problem you posted in that it’s a word problem that yields a system of algebraic equations which, when solved, yields the correct answer. Unlike your problem though, this requires some abstract deductive reasoning, hence it being a logic puzzle, in order to arrive at the equations you need to solve.

[Guest]

BOGUS!!!

[DarthMaximus]

@MADDOGG:

BOGUS!!!

LMAO!!!

[Bashir]

@DarthMaximus:

Now Chuck looks and also has two options, A+B or A-B, but can’t decide, however, this does give Alex the required information he needs to eliminate either B+C or B-C, thus he knows he has 50.

He can eliminate B+C, because if A=B+C => B and C are both 25… He would have known immidietly…
So Alex knows it is B-C

So when Bob sees the hats the first time he sees 50 and C… He can’t make up out of that if it is B=50+C or B=50-C
Chuck also can’t make up his own number. He sees 50 and B… Hee can’t make up if it is C=50+B or C =50-B

Need to think about this…

[Avin]

He can eliminate B+C, because if A=B+C => B and C are both 25… He would have known immidietly…
So Alex knows it is B-C

I’m not sure what you’re saying with this. If you are saying that A=B+C necessarly implies that B and C are both 25, that can’t be right: B could be 1 and C could be 49, for instance.

I’ll give you a hint. What you know is this:

• A has a 50 on his hat. We know this from the end of the problem.
• A does not have enough information after looking at B and C’s numbers to figure out his own number (in otherwords, he doesn’t see two of the same number.)
• Knowing that A doesn’t see two of the same number, B still does not have enough information to determine the number of his hat. This means …. ?
• Knowing the conclusions B might have reached from the previous step, C still does not have enough information to determine his own number. This means … ?
• Knowing the conclusions C might have reached from the previous step, A is now able to determine that his number is 50.

[Bashir]

Avin, after I thought about it for a while it was fairly ez…
A=50 B=20 and C=30

My reasoning behid this…
A thinks he has either 10 or 50 => pass
If A is wearing 10, C will see 10 and 20. He thinks he has 10 or 30.
If B saw 2x 10 he would know he wears 20, this does not happen => That means C realizes that his own hat must show 30 if A’s hat was 10.
=> This does not happen, so A knows his hat is not 10. So he know it has to be 50.

I hope this is correct… Otherwise I don’t know and I will find it Boogus too :P

[Bashir]

Here is one from me.

How can three missionaries and three cannibals cross a river two at a time in a canoe if the cannibals must never outnumber the missionaries left on one side, and two cannibals cannot paddle across together?

[Yanny]

How can three missionaries and three cannibals cross a river two at a time in a canoe if the cannibals must never outnumber the missionaries left on one side, and two cannibals cannot paddle across together?

Missionary A + Cannibal A across.  2/2-1/1
Missionary A back. 3/2-0/1
Missionary A + Cannibal B across 2/1-1/2
Cannibal B back. 2/2-1/1 (Cannibals can paddle themselves right?)
Missionary B + Cannibal B across 1/1-2/2
Cannibal B back. 1/2-2/1
Missionary C + Cannibal C across 0/1-3/2.
Missionary C back.
Missionary C + Cannibal B across. 0/0-3/3

[Bashir]

Missionary A + Cannibal B across 2/1-1/2

someone got slaughtered :D

[Yanny]

He never left the boat, I swear :)

Question, because I don’t think that you said this (though it may be implied). Do two cannibals left alone without a missionary eat each other? If not, the problem is quite easy.

[Avin]

Good job, Bashir! That is indeed the correct answer as well as the correct reasoning for it. For a bonus, you could try to show why that is the ONLY answer possible, (you can prove it mathematically) but I’ll count the problem as essentially solved.

For your problem, what do you mean by the statement that two cannibals cannot paddle across together? Can a single cannibal paddle across alone, or can only missionaries paddle?

[Bashir]

I just know if you take different numbers the problem gets too deep… So A can never argue why he has 50…

2 Kannibals can paddle… And 2 kannabals don’t eat each other… And 1 can paddle across, so both miss and Kan. Not leaving the boat is not implied, 2 kannibals and 1 miss is just eating :D So if 2 kan are across and the miss is still in the boat he still got eaten…

By the way, 2 miss can’t converse 1 kannibal… The is no bridge where they can all move together… There is no leak in the boat… I think you get it now…