Logic problem (1 Viewer)

[Yankee23Fan]

Or... you can clone yourself and make your clone take whatever two pills you want from you hand. If he lives then you kill yourself and he lives on as you with all your memories. If he dies, you know what pill to take. It's the Gattica solution.

 

[jon_mx]

Have we done The Pill Problem?

You're on expensive medication that keeps you alive. You must take one pill of medication A and one pill of medication B per day, or else you will die. An overdose will also be fatal, so under no circumstances can you take two of either medication in the same day. The pills are expensive and rationed, so it's not possible to obtain more pills, so you can't let any go to waste.

One day, you're in the bathroom getting ready for your daily dose. You shake out one pill of medication A, but your hand shivers when shaking out medication B and two pills fall into your palm. Unfortunately, both types of pills look the exact same, they're both the same color, size, shape, and weight. You cannot tell which pill is medication A and which two are medication B.

Since the pills are expensive and rationed, you cannot throw them away and start over. You must ensure that you take exactly one A pill and one B pill today. You can't take all 3 pills because doubling up on B is fatal. What do you do?

Put those 3 pills aside in a sandwich baggie for later review and get a new one of each to take for today.

Assume "later review" will not be helpful, so you're only delaying the problem.

Ok, fine. Mr. rules.

If he is really taking one of each pill each day then the B bottle is going to have one pill less in it. So count the pills and figure out which one has 1 more. That will be A. Take a pill out of A so that you have 2 A pills and 2 B pills out. Close the bottles. Cut the first three in half and put the half of each in one pile. Then cut the new pill a in half and add one half to that pile of three. You then have 2 halfs of each pill and therefore 1 of each pill.

Cool.

 

[Hawk Fan]

Answer is still 2/3 years later.

 

[jon_mx]

Answer is still 2/3 years later.

XXX

 

[jon_mx]

I think I have this one worded right:

Two nomads are riding their camels through the desert. They stop to eat. One has five rolls of bread, and the other has three. Just when they're about to start eating, another man shows up who has no food. Being good samaritans, they offer to share, and each of the three men eats an equal amount of bread. The stranger then offers to pay for his food, and gives the two nomads eight coins of equal value.

The first nomad figures he had five rolls to share, and the other man three, so he should take five coins and gives three to the other guy. The second guy disagrees, and argues since both shared their food they should get four coins each.

Unable to come to a resolution, they find a wandering mathematician who tells them how to distribute the coins. How much is each nomad due?

Why would they be due anything. The food was a gift.

 

[Spartans Rule]

I think I have this one worded right:

Two nomads are riding their camels through the desert. They stop to eat. One has five rolls of bread, and the other has three. Just when they're about to start eating, another man shows up who has no food. Being good samaritans, they offer to share, and each of the three men eats an equal amount of bread. The stranger then offers to pay for his food, and gives the two nomads eight coins of equal value.

The first nomad figures he had five rolls to share, and the other man three, so he should take five coins and gives three to the other guy. The second guy disagrees, and argues since both shared their food they should get four coins each.

Unable to come to a resolution, they find a wandering mathematician who tells them how to distribute the coins. How much is each nomad due?

They all ate 2 2/3 rolls of bread. That means the guy with 3 only had 1/3 of a roll to give, and the other guy gave 2 1/3 rolls.

So wanderer guy paid 1 coin for each 1/3 roll of a bread. 7 for 5-roll guy, 1 for 3-roll guy.

 

[Sinn Fein]

Two former college roommates, both logicians, meet at a conference after many years without contact. While catching up, the two eventually get around to discussing their children. The first logician asks the second how many children he has, and what their ages are. The second replies that he has 3 children, but (ever the logician) he will only reveal clues about their ages. The first logician must deduce for himself the ages of the second's children.

"First," says the logician, "the product of my children's ages is 36."

"Second, the sum of their ages is the same as our apartment number in college."

"Third, my oldest child has red hair."

Upon hearing the third clue, the first logician replies at once with the ages of his friend's children. What are they? How do you know?

 

[fred_1_15301]

Two former college roommates, both logicians, meet at a conference after many years without contact. While catching up, the two eventually get around to discussing their children. The first logician asks the second how many children he has, and what their ages are. The second replies that he has 3 children, but (ever the logician) he will only reveal clues about their ages. The first logician must deduce for himself the ages of the second's children.

"First," says the logician, "the product of my children's ages is 36."

"Second, the sum of their ages is the same as our apartment number in college."

"Third, my oldest child has red hair."

Upon hearing the third clue, the first logician replies at once with the ages of his friend's children. What are they? How do you know?

The different combinations of ages could be:

36, 1, 1

18, 2, 1

12, 3, 1

9, 4, 1

9, 2, 2

6, 6, 1

6, 3, 2

4, 3, 3

Am I missing any here? Of those, 661 and 922 are the only 2 combinations that add up to the same number (13). This requirement would then make the 3rd clue important. If there is an "oldest child", the combination would have to be 922. I think.