Five pirates have obtained 100 gold coins and have to divide up the loot. The pirates are all extremely intelligent, treacherous and selfish (especially the captain).
The captain always proposes a distribution of the loot. All pirates vote on the proposal, and if half the crew or more go "Aye", the loot is divided as proposed, as no pirate would be willing to take on the captain without superior force on their side.
If the captain fails to obtain support of at least half his crew (which includes himself), he faces a mutiny, and all pirates will turn against him and make him walk the plank. The pirates start over again with the next senior pirate as captain.
What is the maximum number of coins the captain can keep without risking his life?
ANSWER:
98
The captain says he will take 98 coins, and will give one coin to the third most senior pirate and another coin to the most junior pirate. He then explains his decision in a manner like this...
If there were 2 pirates, pirate 2 being the most senior, he would just vote for himself and that would be 50% of the vote, so he's obviously going to keep all the money for himself.
If there were 3 pirates, pirate 3 has to convince at least one other person to join in his plan. Pirate 3 would take 99 gold coins and give 1 coin to pirate 1. Pirate 1 knows if he does not vote for pirate 3, then he gets nothing, so obviously is going to vote for this plan.
If there were 4 pirates, pirate 4 would give 1 coin to pirate 2, and pirate 2 knows if he does not vote for pirate 4, then he gets nothing, so obviously is going to vote for this plan.
As there are 5 pirates, pirates 1 & 3 had obviously better vote for the captain, or they face choosing nothing or risking death.
Take the three pills, cut them in half so that you have two sets of three half pills, keeping careful not to mix them. Separate the halves into two separate groups (both sets should now contain A, B, B halves) being careful once again not to mix them up. Take another A pill, and cut that in half. Add one of those halves to the first set of three halves (now they should contain A, A, B, B, halves), and the other to the second set. You now have two sets of halves. In each set is two halves of a B pill and two sets of an A pill.
BARTENDER
A mathematician enters a bar and starts chatting with the bartender. The bartender tells him about his three daughters, and when he is asked about their age, he decides to make it a bit more interesting, as he (the bartender) is also very interested in mathematics. He says, “The product of their ages is 72.” The mathematician answers, “OK, but that didn’t help a lot.” — “Then I should tell you that the sum of their ages is equal to the street number of this bar.” The mathematician leaves the bar, returns, and says, “Great, but I still don’t know their age.” The bartender smiles and says, “My youngest daughter really likes strawberry ice cream.” Now the mathematician knows their ages.
How old are the three daughters?
The captain says he will take 98 coins, and will give one coin to the third most senior pirate and another coin to the most junior pirate. He then explains his decision in a manner like this...
If there were 2 pirates, pirate 2 being the most senior, he would just vote for himself and that would be 50% of the vote, so he's obviously going to keep all the money for himself.
If there were 3 pirates, pirate 3 has to convince at least one other person to join in his plan. Pirate 3 would take 99 gold coins and give 1 coin to pirate 1. Pirate 1 knows if he does not vote for pirate 3, then he gets nothing, so obviously is going to vote for this plan.
If there were 4 pirates, pirate 4 would give 1 coin to pirate 2, and pirate 2 knows if he does not vote for pirate 4, then he gets nothing, so obviously is going to vote for this plan.
As there are 5 pirates, pirates 1 & 3 had obviously better vote for the captain, or they face choosing nothing or risking death.
DUMBASS M.D.
You have Some Terminal Disease, which necessitates taking two pills a day: one Pill A and one Pill B. If you neglect to take either pill, you die; if you take more than one A or more than one B, you die. If you don't take them at exactly the same time, you die.
This morning you are going through your usual routine. You pick up your bottle of A Pills and gently tap one into your palm. Then you pick up your bottle of B Pills and tap it, but two pills accidentally fall into your hand. You now hold three pills (one A and two Bs), you don't know which are which, and they are completely indistinguishable from each other. The A Pills are the same color as the B Pills, they are the same shape, same size -- they appear identical in every respect. Man, your doctor is a dumbass. But he's a rich dumbass, because he's charging you $1,000,000 a pill! So you dare not throw any away.
Thus, the puzzle: what can you do to ensure that you take only one A Pill and only one B Pill today, without wasting any pills (either today or in the future)?
ANSWER:
You have Some Terminal Disease, which necessitates taking two pills a day: one Pill A and one Pill B. If you neglect to take either pill, you die; if you take more than one A or more than one B, you die. If you don't take them at exactly the same time, you die.
This morning you are going through your usual routine. You pick up your bottle of A Pills and gently tap one into your palm. Then you pick up your bottle of B Pills and tap it, but two pills accidentally fall into your hand. You now hold three pills (one A and two Bs), you don't know which are which, and they are completely indistinguishable from each other. The A Pills are the same color as the B Pills, they are the same shape, same size -- they appear identical in every respect. Man, your doctor is a dumbass. But he's a rich dumbass, because he's charging you $1,000,000 a pill! So you dare not throw any away.
Thus, the puzzle: what can you do to ensure that you take only one A Pill and only one B Pill today, without wasting any pills (either today or in the future)?
ANSWER:
BARTENDER
A mathematician enters a bar and starts chatting with the bartender. The bartender tells him about his three daughters, and when he is asked about their age, he decides to make it a bit more interesting, as he (the bartender) is also very interested in mathematics. He says, “The product of their ages is 72.” The mathematician answers, “OK, but that didn’t help a lot.” — “Then I should tell you that the sum of their ages is equal to the street number of this bar.” The mathematician leaves the bar, returns, and says, “Great, but I still don’t know their age.” The bartender smiles and says, “My youngest daughter really likes strawberry ice cream.” Now the mathematician knows their ages.
How old are the three daughters?
ANSWER:
2,6,6
If you break 72 down into its prime factors (1,2,2,2,3,3), it helps to list all of the three number combinations whose product is 72 (also noted below are their sums):
1,1,72 = 74
2,3,12 = 17
1,2,36 = 39
2,4,9 = 15
1,4,18 = 23
2,6,6 = 14
1,6,12 = 19
3,3,8 = 14
1,8,9 = 18
3,4,6 = 13
2,2,18 = 22
If the guest had looked at the house number and said he knew the solution, then the set of ages could have been any of those with unique sums shown above (for instance, if he saw that the house number was 15, then he would have known the ages were 2, 4, and 9, since those are the *only* numbers whose product is 72 and whose sum is 15). But since he said he didn't have enough information, the house number must have been 14 (since there are two sets of numbers which add to 14). The last clue, "the youngest likes strawberry icecream" provides enough information to solve the problem since there is no youngest daughter in the set of numbers 3,3,8. =)
If you break 72 down into its prime factors (1,2,2,2,3,3), it helps to list all of the three number combinations whose product is 72 (also noted below are their sums):
1,1,72 = 74
2,3,12 = 17
1,2,36 = 39
2,4,9 = 15
1,4,18 = 23
2,6,6 = 14
1,6,12 = 19
3,3,8 = 14
1,8,9 = 18
3,4,6 = 13
2,2,18 = 22
If the guest had looked at the house number and said he knew the solution, then the set of ages could have been any of those with unique sums shown above (for instance, if he saw that the house number was 15, then he would have known the ages were 2, 4, and 9, since those are the *only* numbers whose product is 72 and whose sum is 15). But since he said he didn't have enough information, the house number must have been 14 (since there are two sets of numbers which add to 14). The last clue, "the youngest likes strawberry icecream" provides enough information to solve the problem since there is no youngest daughter in the set of numbers 3,3,8. =)
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