Gogf
Indescribable
I guess he could take 96, or possibly even all of it. Most of the crew wouldn't mutiny, because they would realize that that would begin a cycle leading to their own death.
Puzzles,conundrums, riddles and thoughts?
- Thread starter Sidhe
- Start date
98? he gives 1 coin each to 2 to get their vote, and they figure they would rather have 1 coin than be dead . . .
Gogf
Indescribable
97 would also work. That way he would buy a majority of the votes, rather than deadlocking the rest of the crew.
EDIT: Nevermind, didn't see that the captain can vote.
Hitti-Litti
Deity
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- Sep 8, 2006
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- 3,767
Let's see...
5 pirates have a hierarchy: I think that the captain has 2 votes instead of one, when the gang has 6 votes
Everybody wants as much gold they can possibly get. But the pirates are wise, so captain, pirate number 1, won't make a ridiculous suggestion. Because they are wise the next captain, pirate number 2, will vote against captain to get the status of captain for himself. So in this case the voting will go like this:
Yay gets 2+0-3 votes.
Nay gets 1+0-3 votes.
Yay needs only one vote to win, which he can achieve by giving himself 50 gold and some other pirate 50 gold. Cappy and the other gold-receiver will vote Yay=3 votes=50% of votes.
The answer: 50 gold pieces.
5 pirates have a hierarchy: I think that the captain has 2 votes instead of one, when the gang has 6 votes
Everybody wants as much gold they can possibly get. But the pirates are wise, so captain, pirate number 1, won't make a ridiculous suggestion. Because they are wise the next captain, pirate number 2, will vote against captain to get the status of captain for himself. So in this case the voting will go like this:
Yay gets 2+0-3 votes.
Nay gets 1+0-3 votes.
Yay needs only one vote to win, which he can achieve by giving himself 50 gold and some other pirate 50 gold. Cappy and the other gold-receiver will vote Yay=3 votes=50% of votes.
The answer: 50 gold pieces.
That makes 2, but I don't have a riddle right now. Maybe later.
Okay, the bottom two?
Okay, the bottom two?
sorry for keeping you guys in suspense, I was out of the house. Anyway, pboily and taliesin nailed it anyway.
if one monk had the disease only, he would see no other monks with a red dot on his forehead, so he would know he had the disease and thus die after the first meeting.
If two monks contracted the disease, both would first assume that the one monk with the red spot would die of the disease after the first meeting, see him come in for lunch though and know themselves to be ill.
So this continues on for a while, the monks don't show up on the morning of the elventh day, meaning that the monks died after dinner on the tenth day and thus after 30 meetings.
So 30 monks died.
edit: just saw i didn't word it quite s intended, so its 33 monks...
if one monk had the disease only, he would see no other monks with a red dot on his forehead, so he would know he had the disease and thus die after the first meeting.
If two monks contracted the disease, both would first assume that the one monk with the red spot would die of the disease after the first meeting, see him come in for lunch though and know themselves to be ill.
So this continues on for a while, the monks don't show up on the morning of the elventh day, meaning that the monks died after dinner on the tenth day and thus after 30 meetings.
So 30 monks died.
edit: just saw i didn't word it quite s intended, so its 33 monks...
Anyway, the solution to mine goes like this:
If there were two pirates, the captain would keep 100 coins.
If there were three pirates, the captain would give one coin to the junior pirate; if the junior pirate votes against it, the captain is killed, leaving two pirates, and no coins at all for the junior pirate. Ergo the junior pirate supports it.
If there were four pirates, the captain would give one coin to the third-ranked, who would get nothing if the plan failed and the captain died. (Here the junior pirate would get one if the plan failed, and the captain can't be sure he'll vote for the plan without giving him two.)
Following similar logic, in our case the captain gives a coin each to the third- and fifth-ranked pirates, who would get nothing were the plan to fail.
The floor's open, it appears.
If there were two pirates, the captain would keep 100 coins.
If there were three pirates, the captain would give one coin to the junior pirate; if the junior pirate votes against it, the captain is killed, leaving two pirates, and no coins at all for the junior pirate. Ergo the junior pirate supports it.
If there were four pirates, the captain would give one coin to the third-ranked, who would get nothing if the plan failed and the captain died. (Here the junior pirate would get one if the plan failed, and the captain can't be sure he'll vote for the plan without giving him two.)
Following similar logic, in our case the captain gives a coin each to the third- and fifth-ranked pirates, who would get nothing were the plan to fail.
The floor's open, it appears.
I think that means you put another one up.
Well, it's Eran's turn, but he's said he's not posting anything for now. In the meantime, a nice easy one:
You have eight billiard balls, one of which is a bit heavier than the others. Using a simple balance scale no more than twice, how can you find the heavy ball?
You have eight billiard balls, one of which is a bit heavier than the others. Using a simple balance scale no more than twice, how can you find the heavy ball?
Gogf
Indescribable
- Divide the balls into three groups. One group has two billiard balls, and the other two have three.
- Weight the two sets of three balls against one another.
- If they weigh the same, weigh the two remaining billiard balls against one another to find the heavy one.
- If one group of three is heavier than the other, take two of the balls it contains and weigh them against one another. If one is heavier, it is the heavy ball. If they both weigh the same, the third ball is the heavy one.
- Weight the two sets of three balls against one another.
- If they weigh the same, weigh the two remaining billiard balls against one another to find the heavy one.
- If one group of three is heavier than the other, take two of the balls it contains and weigh them against one another. If one is heavier, it is the heavy ball. If they both weigh the same, the third ball is the heavy one.
Irish Caesar
Yellow Jacket
Spoiler :
Edit: It would appear that I am a slow typer, as when I started, there was no response. By the time I finished, Gogf had posted the answer and Taliesin already affirmed it. Ah, well.
Gogf
Indescribable
I don't have one, so the floor is open. Irish Caesar gets priority if he wants it.
I need an exuse for not learning for my exam tomorrow, so i'll post a nice and short one.
You're sitting round on a boat on a (very small) lake. On your boat you have a large chunk of lead. You decide to throw the lead over-board.
What happens with the water level of the lake?
You're sitting round on a boat on a (very small) lake. On your boat you have a large chunk of lead. You decide to throw the lead over-board.
What happens with the water level of the lake?
Nothing or practically nothing.
When it hits the bottom it sinks into the mud leaving the water level pretty much unchanged.
Gogf
Indescribable
Sidhe's right. The lead is already displacing water while it's in the boat.
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