The Riddle Thread….
Home › Forums › Decaffeinated Coffee › The Riddle Thread….
 This topic has 2,250 replies, 179 voices, and was last updated 2 years, 1 month ago by noitallmr.

AuthorPosts

October 22, 2009 6:07 pm at 6:07 pm #1068772YW Moderator80Member
I you have already made the assumption that the fuse burns consistently throughout it’s length, as you need to assume for your answer, then just take one fuse and measure 3/4 of it, and burn it to that point.
October 22, 2009 6:08 pm at 6:08 pm #1068773Dr. PepperParticipantYW Moderator80
I answered that without reading the question carefully because I had to run to Mincha. I realized later that the answer is incorrect.
October 22, 2009 6:10 pm at 6:10 pm #1068774hiijackerParticipantYour 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.
October 22, 2009 6:21 pm at 6:21 pm #1068775Dr. PepperParticipanthiijacker
Make two piles, one with 80 quarters, one with 20.
The pile with 80 quarters will have (t) tails and (80 – t) heads. The pile with 20 quarters will have (20 – t) tails and (t) heads. (0 <= t <= 20)
Flip over every quarter in the pile with 20 quarters and give me $25.
(The pile with 20 quarters will now have (20 – t) heads and (t) tails. (0 <= t <= 20))
Thanks
(Would you like my home address or is Paypal easier?)
October 22, 2009 6:39 pm at 6:39 pm #1068776YW Moderator80Memberlight one fuse from both ends
light the other fuse from one end at the same time
when first fuse finishes (1/2 hour later) then light the other end of the second fuse (which will now burn for 15 more minutes (i think)
October 22, 2009 7:32 pm at 7:32 pm #1068777I can only tryMemberAssumptions:
1) Zero is not an option
2) Two identical numbers aren’t options:
Mr. Product: I do not know the numbers
The sum of the numbers can’t be a prime number, or 2.
Mr. Sum: I knew you didn’t know the numbers
No permutation of numbers totaling the sum of the two numbers will result in a unique product.
I.e. if Mr. Sum’s number was 56, then Mr. Product’s number could have been 159 (a unique 3 * 53 combination).
Mr. Product: Now I know the numbers
By Mr. Sum’s revealing that at least one possible sum’s permutation’s have no unique results, Mr. Product eliminates all numbers whose sum’s may have unique permutations, leaving only a single “no unique” possibility.
The sums whose every possible combinations of numbers result in a product whose product is shared by another two numbers are: 7, 9, 11, 13, 15, 16, 17, 19, 21, 22, 23, 25, 27, 29, 31, 35, 37, 41, 43, 45, 47, 49, 53
Now, the application of bruteforce
Let’s say the numbers are 1 and 6.
The sum is 7, the product is 6.
For Mr. Sum, the numbers can be 1:6, 2:5, 3:4.
For Mr. Product, the numbers can be 1:6, 2:3.
Mr. Sum relizes that Mr. Product can’t know the solution since he may either have 6 (1*6 or 2*3), 10(1*10 or 2*5) or 12 (1*12 or 3*4) as his choices.
Once Mr. Sum reveals that he knew Mr. Product doesn’t know the answer, he eliminates 2:3 as a possibility (since 2:3’s sum is 5, and a sum of five may have resulted from 1:4, and with 1:4 Mr. Product would have known the answer).
Mr. Product has removed 2:3 from the possibilities, leaving only 1:6, so he knows the answer.
If Mr. Product’s number was 10, he couldn’t have eliminated anything, since both 7(1*6) and 11(1*10) are on the above list.
If Mr. Product’s number was 12, he also couldn’t have eliminated anything, since both 7(1*6) and 13(1*12) are on the above list.
Therefore, Mr. Sum relized that Mr. Product eliminated 2*3, leaving 1 and 6.
now I’ll read the resident maven’s solution
October 22, 2009 8:00 pm at 8:00 pm #1068778hiijackerParticipantI can only try, Nice explanation as well.
I have hundreds more of these riddles, I prefer the logic ones, but here is a nonlogic 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?
October 22, 2009 8:06 pm at 8:06 pm #1068779mepalMemberhiijacker: Why do you have two ‘I’s in your SN?
October 22, 2009 8:09 pm at 8:09 pm #1068780I can only tryMember“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?”
Break three in half, leaving each person with an odd number of wholes (49) as well as an odd number of halves (3)
October 22, 2009 8:25 pm at 8:25 pm #1068781hiijackerParticipantNice try, but all the cubes are whole, none of them are cut in half.
October 22, 2009 8:28 pm at 8:28 pm #1068782YW Moderator80Memberput one in one cup and 100 in the other (assuming it’s a very big cup)
October 22, 2009 8:30 pm at 8:30 pm #1068783truthsharerMemberGive 50 to one and 51 to the other. 50 cubes for a cup of coffee is odd.
October 22, 2009 8:41 pm at 8:41 pm #1068784YW Moderator80MemberAlthough I was first, I think ts’s answer is more practical. I concede to him.
To those who think this riddle is unfair, he did say it was a nonlogical riddle.
October 22, 2009 8:44 pm at 8:44 pm #1068785Dr. PepperParticipantPut 8 small cubes together to make one larger cube. You now have 93 small cubes and 1 large cube or 94 all together. Each person gets 47 cubes which is an odd number.
QED
October 22, 2009 8:47 pm at 8:47 pm #1068786YW Moderator80MemberDr. Pepper that can’t be. That is logical and hiijacker specifically said: “but here is a nonlogic riddle”
My answer is definitely less logical than yours.
October 22, 2009 8:47 pm at 8:47 pm #1068787hiijackerParticipanttruthsharer 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??
October 22, 2009 9:08 pm at 9:08 pm #1068788I can only tryMemberFrom “hiijacker”s sum/product puzzle:
“There are two unknown whole numbers, x and y, both greater than 1, and less than 100.
Sigh.
Back to remedial reading 101.
October 22, 2009 9:43 pm at 9:43 pm #1068789truthsharerMemberA woman has two children, one of which is a girl. What is the probability that the other child is also a girl?
October 22, 2009 9:50 pm at 9:50 pm #1068790truthsharerMemberI have to go soon, so can I post the answer now, and then have it approved once a few people attempt an answer?
October 23, 2009 12:30 am at 12:30 am #1068791truthsharerMemberWow, no takers?
October 23, 2009 12:55 am at 12:55 am #1068792Mezonos MavenMember50%.
The gender of the one child who is a girl has no bearing on the probability of the second child also being a girl.
October 23, 2009 1:04 am at 1:04 am #1068793I can only tryMemberA woman has two children, one of which is a girl. What is the probability that the other child is also a girl?
0%
October 23, 2009 1:09 am at 1:09 am #1068794truthsharerMemberIncorrect.
October 23, 2009 1:16 am at 1:16 am #1068795truthsharerMemberIncorrect to both MM and ICOT.
October 23, 2009 1:53 am at 1:53 am #1068796Dr. PepperParticipanttruthsharer
1/3
Do you want me to explain?
October 23, 2009 2:26 am at 2:26 am #1068797truthsharerMemberDr. P. has it right. You can explain if you want to, or we can wait and see if people can figure out why, once they know the answer.
October 23, 2009 3:22 am at 3:22 am #1068798Dr. PepperParticipantJust so no one loses any sleep over this.
A family has two children, there are 4 possible scenarios:
Boy Boy
Boy Girl
Girl Boy
Girl Girl
Since one of the children is a girl there are only three possible scenarios:
Boy Girl
Girl Boy
Girl Girl
Of those three scenarios only one has the desired outcome.
Therefore the probability is 1/3.
QED
(I remember this problem from a probability course. There is actually a massive machlokes between some really great math nerds, I’m not saying what side I’m on, but I knew that if truthsharer was asking this question he was probably not looking for the obvious answer.)
See you all in the morning.
October 23, 2009 3:25 am at 3:25 am #1068799hiijackerParticipantHere 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?
October 23, 2009 3:28 am at 3:28 am #1068800hiijackerParticipantWe 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.”
October 23, 2009 3:36 am at 3:36 am #1068801hiijackerParticipantHere 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?
October 23, 2009 4:30 pm at 4:30 pm #1068802squeakParticipanthiijacker – easy one. The king himself is putting on the hats? In that case, the first prisoner says, “My hat could be black or it could be white. I don’t know and I don’t care because what really matters is THIS MACE THAT I’VE BEEN HOLDING BEHIND MY BACK”. After that, all the prisoners are free to go.
October 23, 2009 5:46 pm at 5:46 pm #1068803neatfreakMemberare they standing in a line where each person sees the person in front of him or are they standing shoulder to shoulder
October 23, 2009 5:54 pm at 5:54 pm #1068804neatfreakMemberoh i see you did say they are facing forward. then they can do something like if its a black hat put your right hand on the shoulder of the person in front of you. if its white put your left hand…
but then i dunno about the last person…
October 23, 2009 5:57 pm at 5:57 pm #1068805neatfreakMemberoh actually if one can guess incorrectly then it would be the last one. so thats all right and they are free.
i kept misreading what you wrote i guess.
October 23, 2009 7:00 pm at 7:00 pm #1068806hiijackerParticipantUmm no. That is not the answer. You cannot signal not kill the king.
October 27, 2009 9:52 pm at 9:52 pm #1068807hiijackerParticipantEveryone stumped?
October 27, 2009 10:04 pm at 10:04 pm #1068808YW Moderator80MemberWe can arrive at this by considering some of the simpler cases. Let’s assume that the king placed only 1 white hat on my head and 99 black hats on everyone else’s head. So at the time when the warden said start, I would look around and see 99 black hats. Since there has to be at least one 1 hat, I would immediately announce that I had a white hat. Everyone else would then know that they had black hats and would say so.
Now consider the case where I have a white hat and someone else also has a white hat. I would look and see 1 white hat. If I had a black hat, that person would immediately announce that they had a white hat. But that doesn’t happen. Instead they wait. Now since neither of us was able to announce right away, we know that there must be 2 white hats. So we both announce that we have white hats.
You can follow this pattern with 3 white hats, etc. but it requires that you have a plan in place for timing. Let’s say that everyone will wait ten seconds for every hat they see less of. So if I see no white hats, I will immediately announce “WHITE” after 0 seconds. Everyone else will follow with “BLACK” right after. Similarly, if I saw 1 white hat, I would plan to say WHITE after 10 seconds. If no one announced after 0 seconds, I would be correct. Then everyone else would say “BLACK” after that.
It doesn’t matter the number of hats or whether there are more or less of one color as long as we wait a discrete period of time to announce. I chose periods of 10 seconds so that there wouldn’t be a chance of mistakes.
If there are W white hats and B black hats, then:
Everyone wearing a white hat will see W1 white hats and B black hats.
Everyone wearing a black hat will see B1 black hats and W white hats.
There are 3 cases:
If W < B then all white hats will say ‘White’ at the same time (10(W1) seconds), and everyone else knows they are wearing black.
If B < W then all black hats will say ‘Black’ at the same time (10(B1) seconds), and everyone else knows they are wearing white.
If W = B then all prisoners will know their color at the same time. (10(B1) seconds, or 10(W1) seconds, they are equivalent).
Let’s take an example. If there were 51 white hats and 49 black hats, and I was wearing a white hat. Then I would plan to say “black” after 490 seconds. However, the people in black hats would beat me to the punch saying “black” after 480 seconds. Then I would say “white” along with all the rest of the prisoners wearing white hats.
October 28, 2009 6:45 pm at 6:45 pm #1068809neatfreakMemberi dont think that will work becasue it said they can only see the hats in front of them.
October 29, 2009 5:25 am at 5:25 am #1068810hiijackerParticipantLet me know when you give up.
October 29, 2009 12:46 pm at 12:46 pm #1068811I can only tryMemberYW Moderator80
Even though your answer doesn’t work with the described setup, it was very lomdish.
hiijacker
You should give a hint or two, like:
1) Remember, every person can see all of the hats in front of him. The last person sees 99, the nexttolast sees 98, etc.
2) We can say with certainty that the last person (the one who can see 99 hats) will see an odd number of…
October 29, 2009 2:15 pm at 2:15 pm #1068812YW Moderator80MemberWell, I guess my google skills need honing. (I only google the riddles when it appears that no one is going to answer it)
October 29, 2009 3:41 pm at 3:41 pm #1068813hiijackerParticipantOk. 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.
October 29, 2009 8:12 pm at 8:12 pm #1068814hiijackerParticipantOkay 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?
October 29, 2009 8:20 pm at 8:20 pm #1068815neatfreakMembermeaning that every mile he needs another banana or that he can first eat all teh needed bananas and then go?
October 30, 2009 12:53 am at 12:53 am #1068816hiijackerParticipantI 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.
October 30, 2009 11:59 am at 11:59 am #1068817I can only tryMemberhiijacker
If all bananas can be eaten before the trip starts, let it eat 1,000 and then carry another 1,000 on the trip. The third 1,000 is (I think) a lost cause. That’s because there’s no way to return and get them.
If a banana has to be eaten before each mile it’s trickier.
Let’s break it into steps:
1) Start out with a load of 1,000
2) Proceed 200 miles, at which point there will be 800 left.
3) Set out back to the starting point with 200 bananas for eating on the trip, leaving 600 at the 200 mile point.
4) Repeat steps 1 – 3. You will now be at the starting point with the last 1,000. 1,200 will be at the 200mile point.
5) Repeat step 1. You will now be at the 200mile point with 2,000 bananas.
6) I’ll leave the rest for someone else.
November 1, 2009 2:16 am at 2:16 am #1068818hiijackerParticipantOk. 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?
November 1, 2009 3:24 am at 3:24 am #1068819I can only tryMemberhiijacker
Full solution, with explanation and numbers for each step:
Until we get down to 2,000 bananas, five bananas will be consumed for each mile traveled, since each mile requires three trips forward and two back. This also gives up 200 total miles traveled from the starting point (1,000 miles divided by five trips).
We are now down to 2,000 bananas remaining, which means we only need three miles traveled for each mile of actual progress. We will divide the 1,000 bananas by three (for the number of trips needed until there are only 1,000 bananas left).
This results in 333 (bananas or miles).
The camel will travel another 333 miles (to the 533 milemark) with a load of 1,000 bananas, return with another 333 that it eats on the way, then pick up the last 1,000 bananas and return to the 533 mile mark while eating another 333.
The camel is now at the 533 mile marker with 1,001 bananas.
The camel is loaded with 1,000 bananas and fed the remaining banana.
It then proceeds the last 467 miles, eating one additional banana for each mile after the first one.
He gets to marked with 534 bananas, and one very wellfed camel.
December 15, 2009 7:50 am at 7:50 am #1068820aussieboyParticipantWhy is this camel eating bananas?! its not a monkey!
December 16, 2009 4:32 pm at 4:32 pm #1068821I can only tryMemberAn oldie but goodie. This is at least 7 – 8 years old.
INSTRUCTIONS
Each equation contains the initials of words that will make it a correct statement. Your job is to finish the missing words. For example: 26 = L. of the A. would be 26 = Letters of the Alphabet. Good luck.
1. 7 = W. of the A.W.
2. 1001 = A. N.
3. 12 = S. of the Z.
4. 54 = C. in a D. (with the j.)
5. 9 = P. in the S. S.
6. 88 = P. K.
7. 13 = S. on the A. F.
8. 32 = D. F. at which W.F.
9. 90 = D. in a R. A.
10. 99 = B. of B. on the W.
11. 18 = H. on a G. C.
12. 8 = S. on a S. S.
13. 3 = B. M. (S. H. T. R.)
14. 4 = Q. in a G.
15. l = W. on a U.
16. 5 = D. in a Z. C.
17. 24 = H. in a D.
18. 57 = H. V.
19. 11 = P. on a F. T.
20. 1000 = W. that a P. is W.
21. 29 = D. in F. in a L. Y.
22. 64 = S. on a C.
23. 40 = D. and N. of the G.F.
24. 2 = T. T.
25. 76 = T. in a B. P.
26. 8 = G. T. in a L. B. C.
27. 101 = D.
28. 23 = S.
29. 4 = H. a J. G. F.
30. 16 = M. on a D. M. C.
31. 5 = G. L.
32. 7 = D. S.
33. 2.5 = C. in a T.A.F.
34. 1, 2, 3 = S. Y. O. at the O. B. G.
35. 3 = M. in a T.
36. 13 = B. D.

AuthorPosts
 You must be logged in to reply to this topic.