# hiijacker

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Viewing 20 posts - 1 through 20 (of 20 total)
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• in reply to: The Riddle Thread…. #1068818
hiijacker
Participant

Ok. After reading your answer, it becomes obvious, the camel cannot eat all bananas before proceeding. The camel eats 1 banana at a time prior to each mile it travels.

So your solution did not provide me with a number. (and the number you will end up with is still not the highest number in any event).

Again, the question is if you start with 3,000 bananas, how many will you be able to get across the 1,000 mile journey?

in reply to: The Riddle Thread…. #1068816
hiijacker
Participant

I solved it understanding that before each mile it travels, it must eat one banana, but I don’t see why he could not eat the total amount needed up front.

in reply to: The Riddle Thread…. #1068814
hiijacker
Participant

Okay here is the answer:

Time for a new riddle:

A banana grower wants to transport his 3000 bananas 1000 miles across the desert to the market. All he can use is his one camel which can carry 1000 bananas at once but it needs to eat one banana to walk one mile (regardless of how many bananas it is carrying).

How many bananas can the grower get to market?

in reply to: The Riddle Thread…. #1068813
hiijacker
Participant

Ok. Here is some information to get you started:

Hmm, he could say the color of the hat on the guy in front of him. That guy would then guess correctly, but then the next guy would be in the same situation as the first guy. Repeating this idea gets only 50 prisoners out guaranteed, with an average of 75 getting out.

But, we can do better. We can actually get the rest of the 99 in line to get ti right.

Please note the answer does not involve any tricks of signaling the answer by touching, or pausing. It only relies on each person guessing black or white.

in reply to: The Riddle Thread…. #1068810
hiijacker
Participant

Let me know when you give up.

in reply to: The Riddle Thread…. #1068807
hiijacker
Participant

Everyone stumped?

in reply to: Is it Muter to Listen to Black Hatitude? #663778
hiijacker
Participant

They both were rabbeim at the time when they made the album. They did not make the album and then do “teshuva”.

in reply to: The Riddle Thread…. #1068806
hiijacker
Participant

Umm no. That is not the answer. You cannot signal not kill the king.

in reply to: The Riddle Thread…. #1068801
hiijacker
Participant

Here is an easy one for the beginners:

A king decides to give 100 of his prisoners a test. If they pass, they can go free. Otherwise, the king will execute all of them. The test goes as follows: t=The prisoners stand in a line, all facing forward. The king puts either a black or a white hat on each prisoner. The prisoners can only see the colors of the hats in front of them. Then, in any order they want, each one guesses the color of the hat on their head. Other than that, the prisoners cannot speak. To pass, no more than one of them may guess incorrectly. If they can agree on their strategy beforehand, how can they be assured that they will survive?

in reply to: The Riddle Thread…. #1068800
hiijacker
Participant

We must assume that the man is telling the truth. To solve this problem we need to know the probabilities of what the man says depending on the type of family he has. Suppose that if he has two boys, then he would say that he has at least one boy with the probability a. Also, suppose that if he has a boy and a girl then he would say that he has at least one boy with the probability b. Then the probability that he has two boys is a/(a+b). There can be several different assumptions:

The case is symmetric towards boys and girls and the man always constructs an answer of the type: “I have at least one boy/girl”. In other words, when he has a boy and a girl, with probability one half, he says that at least one child is a boy, and with probability one half, he says that at least one child is a girl. When he has two boys, he says that he has at least one boy with probability one. In this case the answer is 1/2.

We are interested in discussing boys, and a man who has at least one boy always says that he has at least one boy. In this case the answer is 1/3.

The man wants to have boys and boasts about them as much as possible. In the case that he has one boy and one girl, he says that he has at least one boy with probability one. If he has two boys, he always says that he has two boys. In this case the answer is 0.

We may invent any distribution of probabilities here.

The correct answer is that the problem is undefined. For example, there is no indication in the problem that all fathers with at least one boy will tell you, “I have at least one boy.”

in reply to: The Riddle Thread…. #1068799
hiijacker
Participant

Here is a similar riddle. but a different result

A man says, “I have two children; at least one of them is a boy.” What is the probability that the other one is a boy?

in reply to: The Riddle Thread…. #1068787
hiijacker
Participant

truthsharer got the answer.

My answer usually is 1 in the first cup and 100 in the second. I wait for the person to give me a strange look and ask, but 100 is not odd?!!!

I respond, isn’t 100 cubes of sugar an odd amount to put into a cup of coffee??

in reply to: The Riddle Thread…. #1068781
hiijacker
Participant

Nice try, but all the cubes are whole, none of them are cut in half.

in reply to: The Riddle Thread…. #1068778
hiijacker
Participant

I can only try, Nice explanation as well.

I have hundreds more of these riddles, I prefer the logic ones, but here is a non-logic riddle:

How can you divide 101 cubes of sugar between 2 people such that each person adds an odd number of whole sugar cubes to his cup of coffee?

in reply to: The Riddle Thread…. #1068774
hiijacker
Participant

Your solution that I quote below is not correct, since the fuses do not burn at consistent rates. So when you light from the middle and both ends, you may have one half that burns up in 2 minute, while the other half takes 58 minutes (or half of 58 minutes, since it lit from both ends).

You said:

Light one in the middle (or light both ends), it will take one half hour to burn.

When that is finished burning light each end and the middle of the next one and it will take 15 minutes to burn.

That should take 45 minutes.

in reply to: The Riddle Thread…. #1068771
hiijacker
Participant

So it seems I need to stump the Pepper:

Here goes.

You are blindfolded and enter into a room. The room has a table with 100 quarters scattered. Your are told that 20 of these quarters are tails and 80 are heads. You are also told that if you can split the coins into 2 piles where the number of tails is the same in both piles, then you get to keep all of the quarters. You are allowed to move the coins, but you can never tell what state a quarter is currently in (the blindfold prevents you from seeing, and you cannot tell by feeling it).

How do you go about splitting the quarters so that you can win?

in reply to: The Riddle Thread…. #1068770
hiijacker
Participant

Good job, here is the full explanation, with each exchange, and how to derive the possibilities:

P says: I do not know the numbers.

So, we can be sure that the product that P knows, can be analysed in more than one way. We name such a product a “fuzzy product”.

Here is an example ( P = 50 = 5 * 10 = 2 * 25). P can’t really know whether 5, 10 or 2, 25 are the numbers X, Y. But if P was 55, then the numbers X, Y would definitely be 5, 11.

S says: I knew that you didn’t know them. I do not know them either.

From the first part of S’s answer, we understand that S has such a sum that every possible X, Y pair corresponding to this sum gives a fuzzy product.

So the Possible Sums are : 11, 17, 23, 27, 29, 35, 37, 41, 47, 53

So, let’s name those sums that have only fuzzy product solutions, good sums.

One example to make that more clear: If you take any of the above sums, and check all possible combinations which yield that sum, you will find out that all of them have a fuzzy product.

11 = 2+9 = 3+8 = 4+7 = 5+6

* 2*9 = 18 = 3*6 (so, 18 is fuzzy)

* 3*8 = 24 = 4*6 (so, 24 is fuzzy)

* 4*7 = 28 = 2*14 (so, 28 is fuzzy), and…

* 5*6 = 30 = 2*15 (so, 30 is fuzzy), thus, all combinations for 11 give fuzzy products.

So, it is now obvious that S has one of the above 10 sums, or else he wouldn’t be sure that P has a fuzzy number, and thus wouldn’t have said that “I knew that you did not know them”.

P says: But now I know them!

FGS! (For God’s Sake) How did P found those numbers?!

Well, when P heard S saying that he (S) knew that he (P) didn’t know the numbers in the first place, he made the calculations above and found out that S had one of those 10 sums(which produce only fuzzy products).

But thou shall not forget that P, being P, knows P ðŸ™‚ (the product)

So, P knows P and those 10 possible sums. Hence, the only thing he can do now is to solve those 10 equation systems with 2 unknowns and 2 equations and hope that only one of those equation systems has a valid solution.

{ X + Y = S

{ X * Y = P

with S = {11, 17, 23, 27, 29, 35, 37, 41, 47, 53}

and P known (to our friend P)

But we know from P’s answer that he found the numbers, and that means one thing: only one of those equation systems has a valid solution.

S says: Now, I know them too!

Of course, S does not have the privilege to know the product P. The only thing that he knows (except from S ðŸ™‚ are the ten equation systems from which P found the numbers.

Now that the two math geniuses S, P learned the numbers, it is right time that we learn them too.

From the fact that his friend P could find the numbers, he draws the conclusion that only one of those equation systems has a valid solution. So, S is dying to find the product P, which, by the way, let’s name “rare product” implying that it has the following property: only one of all pairs of numbers which have this product, has a good sum. If this sounds complicated, read until clear.

But S says that he found out the numbers. That can only mean that he has a sum that of all possible pairs which yield that sum, only one has a rare product.

What is therefore left for us is to check all of the 10 possible sums to find the one that has a unique pair of numbers which give a rare product.

We are then presented with:

SUM = 17

PRODUCT = 52

which reassures us that 17 is the one and only of all the 10 possible sums that of all pairs of numbers which sum to it (e.g. 5, 12 or 8, 9) only one has a rare product, i.e. a product which of all pairs of numbers which have that product only one has a good sum, i.e. a sum which all pairs of numbers which sum to it have fuzzy products.

So which numbers have a product of 52 and a sum of 17?

They are 4 and 13…

That’s all folks, I hope you had a nice time!

in reply to: The Riddle Thread…. #1068768
hiijacker
Participant

While you are working on that, here is one of my favorite riddles that can be solved in your head.

You have two one-hour fuses: lighting one end of a fuse will cause it to burn down to the other end in exactly one hour’s time. You know nothing else about the fuses; in particular you don’t know how long any segment of a fuse will burn, only that an entire fuse takes one hour. How can you tell when exactly 45 minutes have passed?

in reply to: The Riddle Thread…. #1068766
hiijacker
Participant

This riddle was posted in Scientific American in the 80’s or 90’s. You can solve the first 2/3 of it in your head, but the last third it helps to have a pen and paper.

I solved this by brute force, without any formulas.

in reply to: The Riddle Thread…. #1068758
hiijacker
Participant

Ok, do not attempt this one unless you have a few days time to kill. This riddle took me over 3 days to solve. But you can solve it in pieces.

There are two unknown whole numbers, x and y, both greater than 1, and less than 100. One mathematician, Mr. Product is given the product of these two numbers, while another mathematician, Mr. Sum is given the sum of these two numbers.

The following conversation takes place:

Mr. Product: I do not know the numbers.

Mr. Sum: I knew you didn’t knew the numbers.

Mr. Product: Now I know the numbers

Mr. Sum: Now I know the numbers, too.

What are the numbers?

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