Logic question

@MADDOGG:

BOGUS!!!

LMAO!!!

@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…

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.

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

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?

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

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

someone got slaughtered :D

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.

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?

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…