The Riddle Thread….
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July 29, 2009 12:03 am at 12:03 am #1068720HidingUnderCoverMember
Rachel, can you flip the dice over, ie. can 02 and 20 share a side?
July 29, 2009 2:29 am at 2:29 am #1068721Rachel613MemberYeah sure you just need to get every combo till 31
July 29, 2009 2:50 am at 2:50 am #1068722HidingUnderCoverMemberBut I assume you cannot flip 13 to become 31, because then it looks backwards. (2 kind of looks the same upsidedown)
July 29, 2009 2:57 am at 2:57 am #1068723Rachel613MemberNo i didn’t mean you should flip them over literally, i meant they can switch places. you can move the dice and put any one first or second.
ICOT – thanx for the welcome! I love this riddle. someone told it to me during a long bus ride in camp and it kept me busy for about an hour….till i chapped the main point 🙂 at least i was distracted!
July 29, 2009 3:04 am at 3:04 am #1068724HidingUnderCoverMemberThen I give up! Mercy on me?
July 29, 2009 3:11 am at 3:11 am #1068725Rachel613MemberTell me what you got so far and where you’re stuck…try to put numbers on and you’ll get to one vital glitch. what have you got for me?
August 6, 2009 2:42 am at 2:42 am #1068726Rachel613MemberWow no one could figure this one out since a week ago?!!! It ain’t that hard!
August 6, 2009 4:46 am at 4:46 am #1068727aussieboyParticipantrachel613: I just saw it. Give me a bit of time to figure it out 🙂
August 6, 2009 4:53 am at 4:53 am #1068728aussieboyParticipantThe problem isnt figuring out which numbers to use. Its figuring out which numbers go where.
August 6, 2009 5:04 am at 5:04 am #1068729aussieboyParticipant1st die: 1, 2, 3, 4, 5, 9
2nd die: 6, 11, 13, 19, 25, 29
I feel like I did to much work though and that there is some kind of shortcut or easier answer.
August 6, 2009 5:10 am at 5:10 am #1068730aussieboyParticipantActually you can technically just do this
1st die: 1, 2, 3, 4, 5, 6
2nd die: 7, 14, 21, 28, x, x (x = any number [because it doesnt matter])
I figured this out after like ten minutes but for some reason it didnt make sense to me until I thought about it for another 10 minutes.
August 6, 2009 2:30 pm at 2:30 pm #1068731mepalMemberHey! Look who I see on the board! Nice seeing ya, aussie! Stick around!
August 6, 2009 8:40 pm at 8:40 pm #1068732aussieboyParticipantthanks mepal. Its doubtfulk but maybe 🙂
August 6, 2009 8:43 pm at 8:43 pm #1068733mepalMemberAt least we get a maybe!
August 6, 2009 9:07 pm at 9:07 pm #1068734I can only tryMemberaussieboy
The format the puzzle actually wants is as follows:
??? ???
?0? ?1? <== first of month
??? ???
??? ???
?1? ?1? <== eleventh of month
??? ???
??? ???
?3? ?1? <== thirtyfirst of month
??? ???August 7, 2009 11:22 pm at 11:22 pm #1068735aussieboyParticipant1st die: 1, 2, 3, 4, 5, 6
2nd die: 0, 1, 2, 7, 8, 9
This is correct as long as 1, 2, 3… is good and it does not need to be 01, 02, 03…
Otherwise I dont know how there can be a possible answer.
August 9, 2009 3:02 am at 3:02 am #1068736I can only tryMemberaussieboy
You’re 99% of the way there.
You still need something else for 07, 08 and 09 to be possible.
August 12, 2009 8:18 pm at 8:18 pm #1068737aussieboyParticipantI give up.
August 12, 2009 9:59 pm at 9:59 pm #1068738I can only tryMemberaussieboy
A hint is in the above sentence.
August 12, 2009 10:36 pm at 10:36 pm #1068739aussieboyParticipant1st die: 0, 1, 3, 4, 5, 6
2nd die: 0, 1, 2, 7, 8, 9
Because 5 and 2 look the same when flipped?
August 12, 2009 11:53 pm at 11:53 pm #1068740I can only tryMemberaussieboy
Oooooh, you’re so close.
It isn’t the 2 that gets flipped…
August 13, 2009 4:25 am at 4:25 am #1068741aussieboyParticipant1st die: 0, 1, 2, 3, 4, 5,
2nd die: 0, 1, 2, 7, 8, 9
ICOT: At this point i didnt really solve the riddle lol
August 13, 2009 4:46 am at 4:46 am #1068742JaxMemberaussie: long time no see buddy! ok, i won’t ruin your concentration in this thread!
August 13, 2009 1:32 pm at 1:32 pm #1068743mepalMemberYeah jax, its a good thing this threads around. Otherwise….
August 13, 2009 4:49 pm at 4:49 pm #1068744I can only tryMemberaussieboy
August 14, 2009 3:13 am at 3:13 am #1068745aussieboyParticipantHey Jax, and mepal
ICOT: Its a deal 😉
August 14, 2009 4:57 pm at 4:57 pm #1068746Mezonos MavenMemberOctober 18, 2009 2:01 pm at 2:01 pm #1068747I can only tryMemberChange is good: I
You will play the following game with one other player and a roll of fifty pennies.
Each player must take one to six pennies from the roll on alternating turns.
The player who takes the fiftieth penny loses.
You go first.
How many pennies should you remove on your first turn? Why?
(from a game magazine)
Change is good: II
The director of the U.S. mint challenges you to the following game:
1) No quarter may touch any other quarter.
Should you go first or second?
What should your strategy be?
(from an online puzzle site)
October 18, 2009 2:19 pm at 2:19 pm #1068748I can only tryMemberChange is good: I
Correction: The player who takes the fiftieth penny wins.
October 18, 2009 3:50 pm at 3:50 pm #1068749Dr. PepperParticipantI can only try
OK I gave it an hour and no one answered.
Change is good: I
I would remove 1 on my first turn.
Each subsequent turn I would pick the amount that when added to the amount he took would equal 7. (If he picked 2 I would pick 5…)
Therefore 1 => 8 => 15 => 22 => 29 => 36 => 43.
There are now seven pennies left.
Any number he chooses to remove now will leave as little as one and as much as six left. I choose the rest and win!
October 18, 2009 4:07 pm at 4:07 pm #1068750I can only tryMemberDr. Pepper
Bingo!
I solved it by realizing that you needed to put down the 43rd penny, which meant you needed to put down the 36th penny, etc. This method works, but is cumbersome.
October 19, 2009 2:18 am at 2:18 am #1068751Dr. PepperParticipantI can only try
Change is good: II
I’ve been racking my brain all day on this one but can’t come up with the correct answer. I think this is probably the one you are looking for (although it is wrong).
I would go first.
To decide where to put the first quarter I would draw two straight, nonparallel lines on the table (with my trusty pencil of course) where each end of the line touches the circumference of the table. I’d then rip out my protractor that I always carry in my pocket protector and use it to find the midpoint of each line. Next I would draw a perpendicular bisector for each line towards the other end of the table. The point where the two lines meet is the exact center of the table.
The center of the first quarter is placed on the center of the table.
After Mr. Director places a quarter I would place a quarter on the exact opposite side of the table. Therefore if he has an empty spot to place a quarter there should be an empty spot on the exact opposite side. Hence I will be the last on to place a quarter on the table.
This will not work once the diameter of the table reaches a certain length which I can definitely not calculate while in the state of mind I am currently in. (It has to do with my favorite constant.)
October 19, 2009 11:24 am at 11:24 am #1068752I can only tryMemberDr. Pepper
Fantastic!
You got the solution, and even described how to locate the exact center of the table, rather than just saying “put the first quarter in the middle”.
I didn’t get the answer to the puzzle, but the “chap” was so clever that I stored it in the vault.
As far as I can tell, this will work for any diameter table, except one so small that a second quarter can’t fit. If this is wrong, I’m interested in how/why.
(As an aside, I think one of New York’s professional sports teams is regularly enjoying celebrations which involve the use of your favorite constant – kain yirbu.)
October 20, 2009 1:24 am at 1:24 am #1068753Dr. PepperParticipantI can only try
Take seven Nickels (or any coin for that matter but preferably without ridges), put one in the center and the other six around it. Look closely and you will see a space between two of the coins. That space is twice the decimal of PI (multiplied by a constant).
Let’s say Mr. Director starts a pattern that is a family of concentric circles centered at the center of the table. In each subsequent circle the little space will get bigger and bigger until there is room for another quarter. Once there is an odd amount of quarters in a circle there will be a space for Mr. Director to place a quarter without a corresponding spot on the “other” side of the table.
QED
October 20, 2009 2:40 am at 2:40 am #1068754I can only tryMemberDr. Pepper
If I understand you correctly:
1) Put a nickel on a flat table.
3) If the six outer nickels are all touching the inner one, there is no way that each outer nickel can touch both adjacent ones, although they will come very close.
4) The total gap between the surrounding nickels is a function of pi. (That space is twice the decimal of PI [multiplied by a constant]).
Is the above correct?
Eventually there will be space(s) large enough for a nickel within the circle, and if the circle is completed by an odd number of coins the director will win.
You can ensure that there is never a circle with an odd number of coins or an unbalanced gap.
By putting your coin exactly opposite his, the gap in the circle must remain divided evenly between the two sides.
October 20, 2009 6:47 pm at 6:47 pm #1068755I can only tryMemberAttentionGetting Quiz
2) The only difference in the green, yellow, red and purple variety of these vegetables is their stage of ripeness.
3) Capsaicin is an irritant in this device, used by lawenforcement agencies.
4) 1885 is the year a pharmacist invented this. After 1950 it was still produced, although it no longer had a point.
5) A Scoville scale will tell you this (not your weight).
6) A variety of spiced meat uses these for flavoring (not the seemingly eponymous food).
October 20, 2009 7:25 pm at 7:25 pm #1068756squeakParticipantDr. Pepper
Member
a space between two of the coins. That space is twice the decimal of PI (multiplied by a constant).
Wouldn’t it also be 100 times the decimal of PI, multiplied by a different constant?
October 22, 2009 6:34 am at 6:34 am #1068758hiijackerParticipantOk, 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?
October 22, 2009 2:58 pm at 2:58 pm #1068760Dr. PepperParticipantI haven’t solved it just yet but I think I’m getting there.
Does anyone mind if I give hint?
October 22, 2009 3:05 pm at 3:05 pm #1068761I can only tryMemberDr. Pepper
Does anyone mind if I give hint?
No hints or answers for another hour, please.
October 22, 2009 3:09 pm at 3:09 pm #1068762Dr. PepperParticipantI don’t have the answer yet, I only did as much as I could think of mentally while on the train this morning. The hint I wanted to give is just to notify others that a notsofamous conjecture may be needed.
October 22, 2009 4:03 pm at 4:03 pm #1068763I can only tryMemberStill not there.
One more hour, please.
October 22, 2009 4:06 pm at 4:06 pm #1068764Dr. PepperParticipantOK but it’s taking lots of self control.
(I have the whole answer now.)
October 22, 2009 4:49 pm at 4:49 pm #1068765I can only tryMemberIt’ll take me too long.
Go ahead with the answer, and I won’t look.
October 22, 2009 5:24 pm at 5:24 pm #1068766hiijackerParticipantThis 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.
October 22, 2009 5:25 pm at 5:25 pm #1068767Dr. PepperParticipantHere it goes.
Mr. Product: I do not know the numbers.
=> Either none of the numbers are prime or one of the numbers is prime but not both. (If both numbers are prime then Mr. Product would just factor the product and have the number.)
Mr. Sum: I knew you didn’t knew [sic] the numbers.
=> The sum can’t be even. (Goldbach’s conjecture – Every even integer greater than 2 can be written as the sum of two primes. Therefore the sum can’t be even or it can be written as the sum of two primes which we know the numbers are not two primes. Sorry everyone but the proof is beyond the scope of this thread.)
=> One number must be odd and one number must be even.
=> The product must be even.
Mr. Product: Now I know the numbers
=> There is one unique way to attain that product using an odd number and an even number. (There is at least one other way, possibly more, to attain that product using two even numbers since one number must be composite.)
Mr. Sum: Now I know the numbers, too.
=> All of us in the CR also do!
4 & 13
There aren’t as many possibilities as you may think once you start throwing out possible pairs based on criteria. The upper bound is “red herring” as it will be the same answer regardless on whether it is 15 or 1,000,000. I tried an upper bound of 10 and found no solution. When I raised it to 15 I found this solution which I believe is the only solution regardless of the upper bound.
October 22, 2009 5:27 pm at 5:27 pm #1068768hiijackerParticipantWhile you are working on that, here is one of my favorite riddles that can be solved in your head.
You have two onehour 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?
October 22, 2009 5:36 pm at 5:36 pm #1068769Dr. PepperParticipantLight 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 5:47 pm at 5:47 pm #1068770hiijackerParticipantGood 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 🙂 (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 🙂 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!
October 22, 2009 5:57 pm at 5:57 pm #1068771hiijackerParticipantSo 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?

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