The worlds first riddle!

Try
That stopping off for coffee is a valid answer as long as - usually - they both stop for coffee on the way home but this time she joins him there. It should be possible to figure out how long their coffee breaks last, but having been accused by you of taking overlong coffee breaks I am sensitive on the subject.

I was thinking about you tipping the cube at 60 degrees. The question arising is how you would 'know' it was 60 degrees and still be within Relative's definition of 'measurement'?

Hey- it's not my concept of measurement. Where'd you get that? Measurement is using agreed upon standards with procedures of measurements to obtain physical quantities.
For measuring quantities that apply in the cube problem, it is forbidden to use: meter, liter, square meter, clock.
You don't need a measuring device to determine a 60 deg. angle. This is a so called 'construction with a ruler and compass' and it can only determine some dimensionless proportions, you cannot measure anything with it. In principle, the whole riddle is a problem of 'construction with a ruler and compass' kind.

Strictly speaking, measuring angles is not measuring at all - as long as you don't use reference angles in form of a physical device. You can measure 60 deg. using an equilateral triangle.

And tilting the cube 60 deg. gives 1/4 of a volume of the cube. This can be seen using the mentioned equilateral triangle, half of which is formed from points A, E and the point where water meets the AB edge (cuts it in half).

Relative

I did not intend to imply that your definition of measurement is different or not different from anyone else's. But within the context of that problem, you did set the problem so you defined the terms.

Here's one for the nutcracker:

WARNING!! Very difficult !!

Visiting Tryagain, Iacomus (no relation to characters on this forum) left his motel and walked around the city for a whole day. After a quick snack, he decided to return to the motel. Tryagain suggested he should return via walking the streets that he had already walked on an uneven number of times ONLY. Prove that not only he can try, he is bound to succeed every time!

"I did not intend to imply that your definition of measurement is different or not different from anyone else's."

Well, that's not how I read it. r/e/a/d/i/n/g/ between the lines it looked like he was saying. "You have never surfed a 360 in your life, and would be better off with a skateboard." Of cause, I may be wrong and I thought I was once. However, I was wrong. :wink:

"Tryagain suggested he should return via walking the streets that he had already walked on an uneven number of times ONLY" That Tryagain, is an idiot.

Relative

In view of what you said earlier about 'PMs flying about', it is with some trepidation and a feeling of temerity that I have sent a PM concerning the oddly walked streets.

Try

How little you know me. If anyone says they have surfed a 360 - whatever that means - I would neither believe it nor disbelieve it until there was some evidence. Missouri mules come to me for lessons!

"The Missouri Mule is Missouri's official state animal. They are a special variety of mule, their horse component being small but powerful breeds such as Shetlands. Their exceptional intelligence is often misdiagnosed as stubbornness.

If you tell a horse to jump over a fallen log, the horse will be thinking 'OK, you're the boss' and jump over the log. A mule will stop dead in his tracks and be thinking 'You idiot. Can't you tell there's a ditch on the other side of that log?' That's not stubborn. That's smart."

You are a good teacher.

I had worked out that it would take about 8 days or so until informed by Try that his "cute looking Mule" was, in fact, "DEAD in the centre". This makes me question Try's taste in cute quite a lot but I will leave it be.



John the priest was obviously the only man in the room.



Take one nut off each of the other three wheels.

As for those forgetful Simpsons and the presents... Stuff 'em! If they're too bloody stupid to remember then they don't deserve presents, and besides, what sort of mother gives a pocketknife to their daughter for christmas?!?

Anaswer to old problems.

The Simpsons:
First, we simplify and number the statements.
A1: water colours to Chester
A2: ribbon to Alice
A3: model train from Deirdre
A4: diary from Alice
A5: diary and pocket watch to Deirdre
A6: loudspeakers from Chester to Bruno
A7: textbook from Bruno
A8: pocket knife not from Chester
B1: DVD player to Alice
B2: textbook to Chester
B3: diary and chocolates to Bruno
B4: chocolates from Alice
B5: socks from Bruno
B6: DVD player from Bruno
B7: model train from Bruno
B8: mittens from Alice to Bruno
C1: pocket watch to Bruno
C2: ribbon to Bruno
C3: DVD player from Chester
C4: pocket watch from Alice
C5: ribbon from Chester
C6: diary from Chester
C7: textbook from Chester
C8: pocket knife from Chester to Deirdre
D1: socks to Alice
D2: loudspeakers to Bruno
D3: DVD player to Deirdre
D4: model train to Bruno
D5: water colours from Deirdre to Chester
D6: chocolates to Chester
D7: pocket watch to Chester
D8: textbook and ribbon to Alice
Figuring out which family member is always correct would be helpful. C3, C5, C6, C7 and C8 together claim that Chester gave five gifts. Since each family member only gave one gift to each of the other family members, Chester is not always correct.
B3 and B4 mean that Alice gave Bruno the box of chocolates, but B8 says her gift to him was the mittens. So Bruno is not always correct.
Let's assume Deirdre is always correct. This gives us the following:
Alice Bruno Chester Deirdre
Alice ------
Bruno ------
Chester ------
Deirdre Water colours ------
Socks Loudspeakers Chocolates DVD player
Textbook Model train Pocket watch
Ribbon
T A1: water colours to Chester
T A2: ribbon to Alice
A3: model train from Deirdre
A4: diary from Alice
F A5: diary and pocket watch to Deirdre
A6: loudspeakers from Chester to Bruno
A7: textbook from Bruno
A8: pocket knife not from Chester
F B1: DVD player to Alice
F B2: textbook to Chester
F B3: diary and chocolates to Bruno
B4: chocolates from Alice
B5: socks from Bruno
B6: DVD player from Bruno
F B7: model train from Bruno
B8: mittens from Alice to Bruno
F C1: pocket watch to Bruno
F C2: ribbon to Bruno
C4: pocket watch from Alice
C6: diary from Chester
At least one of C3 and C8 is false:
C3: DVD player from Chester
C8: pocket knife from Chester to Deirdre
At least one of C5 and C7 is false:
C5: ribbon from Chester
C7: textbook from Chester
T D1: socks to Alice
T D2: loudspeakers to Bruno
T D3: DVD player to Deirdre
T D4: model train to Bruno
T D5: water colours from Deirdre to Chester
T D6: chocolates to Chester
T D7: pocket watch to Chester
T D8: textbook and ribbon to Alice
No contradictions so far. We can conclude that Alice has to be the 3/4 correct family member, since Bruno and Chester both have four false statements each. Let's add another assumption: That Bruno is right half the time. That means that B4, B5, B6 and B8 have to be true. B5 means that A7 has to be false (or Bruno gave Alice two presents). Since we now know Alice's two false statements, the remaining one's have to be true.
A4 and A8 mean that C6 and C8 are false. B6 means that C3 is false. B4 means that C4 is false (else Alice gave Chester two presents). But since one of C5 and C7 have to be false (else Chester gives Alice two presents), Chester has at most one true statement, which is less than one quarter.
That's a contradiction, so if Deirdre is always correct, Bruno is only correct one quarter of the time and Chester is correct half the time. Moving back to before we assumed Bruno to be correct half the time, this means that C4 and C6 are true. C6 means that A4 is false, which once again leads us into the situation of knowing that Alice's remaining statements are true. C4 means that B4 is false, A7 means that B5 is false. A8 means that C8 is false so C3 is true, which means that B6 is false. So Bruno has seven false statements, so the assumption that Deirdre is always correct must be false. Therefore, Alice is the one that is always correct.
Jumping back to the very beginning, this gives us the following:
Alice Bruno Chester Deirdre
Alice ------ Diary
Bruno ------ Textbook
Chester Loudspeakers ------
Deirdre ------ Model train
Ribbon Water colours Pocket watch
T A1: water colours to Chester
T A2: ribbon to Alice
T A3: model train from Deirdre
T A4: diary from Alice
T A5: diary and pocket watch to Deirdre
T A6: loudspeakers from Chester to Bruno
T A7: textbook from Bruno
T A8: pocket knife not from Chester
B1: DVD player to Alice
B2: textbook to Chester
F B3: diary and chocolates to Bruno
B4: chocolates from Alice
B5: socks from Bruno
B6: DVD player from Bruno
F B7: model train from Bruno
B8: mittens from Alice to Bruno
F C1: pocket watch to Bruno
F C2: ribbon to Bruno
C3: DVD player from Chester
F C4: pocket watch from Alice
C5: ribbon from Chester
F C6: diary from Chester
F C7: textbook from Chester
F C8: pocket knife from Chester to Deirdre
D1: socks to Alice
T D2: loudspeakers to Bruno
D3: DVD player to Deirdre
D4: model train to Bruno
D5: water colours from Deirdre to Chester
D6: chocolates to Chester
F D7: pocket watch to Chester
D8: textbook and ribbon to Alice
Since Chester has six false statements, the remaining two must be true. So the silk ribbon is from Chester to Alice, and the DVD player is from Chester to Deirdre (since he gave Alice the ribbon and Bruno the loudspeakers). Therefore, B1 and B6 are false, and this gives Bruno four false statements, telling us that B2, B4, B5 and B8 are true.
This gives us:
Alice Bruno Chester Deirdre
Alice ------ Mittens Diary Chocolates
Bruno ------ Textbook Socks
Chester Ribbon Loudspeakers ------ DVD player
Deirdre ------ Model train
Water colours Pocket watch
T A1: water colours to Chester
T A2: ribbon to Alice
T A3: model train from Deirdre
T A4: diary from Alice
T A5: diary and pocket watch to Deirdre
T A6: loudspeakers from Chester to Bruno
T A7: textbook from Bruno
T A8: pocket knife not from Chester
F B1: DVD player to Alice
T B2: textbook to Chester
F B3: diary and chocolates to Bruno
T B4: chocolates from Alice
T B5: socks from Bruno
F B6: DVD player from Bruno
F B7: model train from Bruno
T B8: mittens from Alice to Bruno
F C1: pocket watch to Bruno
F C2: ribbon to Bruno
T C3: DVD player from Chester
F C4: pocket watch from Alice
T C5: ribbon from Chester
F C6: diary from Chester
F C7: textbook from Chester
F C8: pocket knife from Chester to Deirdre
D1: socks to Alice
T D2: loudspeakers to Bruno
D3: DVD player to Deirdre
D4: model train to Bruno
D5: water colours from Deirdre to Chester
D6: chocolates to Chester
F D7: pocket watch to Chester
F D8: textbook and ribbon to Alice
As can be seen from the table, the pocket watch is from Bruno, the socks are Alice's, the box of chocolates belongs to Chester and the water colours are from Deirdre. Also, since D7 and D8 are both false, D4 has to be true, leaving only one spot for the pocket knife; from Deirdre to Alice.

The gambler thought for a moment, then answered correctly. How did the five horses finish the race?

Whispered Promises came in first. Skipper's Gal and Happy Go Lucky tied for second place. Penuche Fudge came in fourth. Near Miss came in fifth.

How many birds:
So five yellow birds were seen (the one Mabel saw, the one Calib saw, the one Abel and Calib saw, the one Mabel and Calib saw, and the one all three saw), and two non-yellow birds were seen (the one Abel saw and the one Abel and Mabel saw) by the group.

Card Machine:
We will first assume the ace was in the second slot after the first shuffle. So the shuffling algorithm would always place the card in the first slot into the second slot. Therefore, the 9 must have been the first card after the first shuffle, or it couldn't have ended up as the second card after the second shuffle.
Now that we know that the 9 must have been the first card after the first shuffle, we know that the shuffling algorithm takes the card in the ninth slot and puts it into the first slot. So after the first shuffle, the 10 must have been in the ninth slot, or the 10 would never have ended up as the first card after the second shuffle.
We follow this logic pattern until our knowledge of the order of the cards after the first shuffle is complete. It turns out that this is the correct answer. If we assume the ace to be in any other position other than the second, then we will eventually encounter a contradiction, where two cards must go into the same slot.
So the final ordering is: 9, A, 4, Q, J, 7, 3, 2, 10, 5, K, 8, 6.


Swimming race:
A won the gold medal; D won the silver medal; C won the bronze medal.

Football.

There are two parts to solving this puzzle. First you need to find the probability of me predicting an individual game correctly.
P(8 winners at least once) = P(not predicting 8 winners all season) = 0.5 = (1-P(correct)^8 not recognized by token OTHER.)^22

Solving for P(correct):

P(correct) = 0.64781

The second part is to find the binomial probability of 6 successes in 8 trials for p = 0.64781, which is 0.25668.

This is the probability of predicting exactly 6 winners in a round. Over 22 rounds, that should happen 5.64703 times.


The letters in Group 1 are homophones of complete English words.

A = a
B = be, bee
C = see, sea
I = I, eye
J = jay
L = ell (measurement of about 45 inches)
M = em (printer's measure)
N = en (another printer's measure)
O = owe
P = pea
Q = cue, queue
R = are
T = tea
U = you, ewe
Y = why

Math:
Complete the following equations with +, -, *, /. Each number must appear in your equations (and only once).
Equations
1 2 3 4 = 28
4*[(2*3)+1] = 28

2 3 4 5 = 28
4*[(2*5]-3] = 28

3 4 5 6 = 28
4*[5+(6/3)] = 28

4 5 6 7 = 28
(4*7)*(7-6) = 28



Scientists
(11 6) = 462 Locks
(462 * 6) / 11 = 252 Keys

Riddle:
In the dripping gloom I see
A creature with broad antlers,
Motionless. It turns its head;
One gleaming eye devours the dark.
I hear it cough and clear its throat;
Then, with a hungry roar, it charges into the night
and is swallowed whole.

Motor bike.

Snow:
At time t, the amount of snow is some constant(r) times t.

snow = rt

The velocity at which the snowplough can travel is inversely proportional to the amount of snow:

v = c/rt

Since the plough travelled the same distance from noon to one as it did from one to four, we can set these distances equal, and use the integral of the velocity as the distance. The snow started at some time x before the plough, so the integrals are from x to 1+x, and from 1+x to 4+x. These integrate to:

r/c(ln(1+x) - ln(x)) = r/c(ln(4+x) - ln(1+x))

The constants cancel, and rearranging use logarithmic properties gives:

ln((4+x)(x)/(1+x)^2) = 0
(4+x)(x)/(1+x)^2 = 1
(4+x)(x) = (1+x)^2

Which simplifies to x=1/2. So, the snow started at 11:30 AM.

That, I hope brings the thread up to date.

As for the visit riddle. All streets would be walked an odd number of times, unless you went out and back and returned to the start point. Therefore, any trip away from the start point would be an uneven number.

A quick word to Adrian. Your restraint is to be applauded.



Today's offering:

My little sister barged into my room clutching a paper bag. "Bet you can't guess how many stickers I bought for $6.25!"
"I guess I won't bother to guess. There's no way anyone could know how many stickers are in that bag," I told her.

"Figure it out," said the little smart aleck. "The stickers were all the same price. The number I bought is the same as the number of cents in each price."

"Hmmm...well, OK ?-now I know it ?- but I'm not going to tell you!" (I hate to admit when I'm stuck.)

(Are you stuck, too?) Think about it: How many stickers are in the bag and how much did each one cost

PM to Try re; stickers.



[Any resemblance to anyone alive or in Georgia is coincidental]

Question #1:

A guy called Relativity wins 3 000 beers in a lottery. The problem is that they are 1 000 miles away and he has to collect them which he is unable to do. If he does not collect them within a stipulated time they are forfeit, and time is running out.

However, a guy called Try-it-on offers to transport them in his pickup truck at a fee of one beer for every mile travelled. His pickup can only hold 1 000 beers at a time, but as the trip is across wasteland he can hide the beers and return for them if required.

As there are no other offers, Relativity has no choice but to accept, but he accepts on the condition that he sets the schedule and itinerary.

Assuming that each of them is honest but intends to have as many beers at the end of this as honesty allows. what conditions does Relativity set out to ensure that he receives as many beers as possible and how many beers would this be?

------------------------------------------------------

Question #2:

[This one can be solved in seconds if one spots the right approach]

Try-it-on and Relativity go out for a beer.

T says, "If we add half of your money to all of my money we can buy a beer each".

R asks with some justified suspicion, "What if we add half of your money to all of my money?"

T says; "then we could by four beers"

How much money, in beers and fractions of a beer, does Try-it-on have?

--------------------------------------------------------------------

Question #3:

I find a piece of old parchment with something written on it, but the writing is too small and faded to be read.

I examine it through a large magnifier which expands everything enormously; the parchment is magnified, marks on the parchment are magnified . . .

Strangely, the writing on the parchment is not magnified.

Any explanations?

PS. Since receiving Adrian's PM I now know that there are two good answers to question 3.

Try, today's offering is not a riddle! That is a grade 2 maths question!

PM'd Iacomus for his three.

A little thought on the streets problem:

Say we have waypoints S,A,B,C,D,E,F with a street between any two waypoints.

You can go SABCDABCDEF and you'll travel AB,BC,CD roads 2 times, but always in the same direction(!).

"Say we have waypoints S,A,B,C,D,E,F with a street between any two waypoints."

I would say. No matter what the layout of streets, unless you return to the start point all movements will be an odd number.
Unless, your wording, "walked around the city for a whole day" He was walking in circles. But, even then once he passed the start point, he would be back to an odd number.

"Since receiving Adrian's PM I now know that there are two good answers to question 3"

I wish I could think of one.

Relative

You are on the ball. The 'odd street' method breaks down when loops are involved. I missed that.

Now that this has been brought to my attention - finally - and to go back over old ground; I think your 'Zings' solution breaks down with an odd number of Zings and that with an odd number of Zings there is no reliable solution.

The long walk home:

The man starts to walk at 14:00. Since the walk saved 10 minutes in the end, those 10 would be spent driving otherwise, meaning 5 minutes in each direction. So when the man met with his wife, it was 14:55 (saved 5 minutes before 15:00). He was walking for 55 minutes.



Sally and Sue:

Sally asks Sam "Is your house number a perfect square?". He answers. Then Sally asks "Is is greater than 50?". Sally thinks she knows the house number now. There are two perfect squares (64,81) between 50 and 100, and Sally herself must live at one of the numbers. Since only the second q. was answered truthfully, Sam's number is not a square, but is greater than 50.
From answers to Sue we know that the Sue's house number is a perfect cube greater than 25, (27 or 64).
Since Sam's number is less than Sally's or Sue's and greater than 50, Sue lives at 64.
We know sum of their numbers is two times a perfect square, the numbers that come to mind are (others are too small or too large):
10*10*2=200
11*11*2=242
If we assume Sue and Sally live both at 64, substracting we get 72 and 114 respectively for Sam's address which is not possible.

If we assume Sue and Sally at 64 and 81, substracting we get 55 and 97 , respectively for Sam's address, meaning 55 is OK and the numbers are:
55(Sam),64(Sue),81(Sally)

Iacomus:


I don't see that. Could you please give more info?

Zings.

Let us call the three 'enmities' x, y, and z. Let x be the enmity B has for A, the Zing who has already been put into a group and y the enmity for the C, the next Zing to be put into a group after B. z is the enmity that may be shared with the Zing in your own group.

Given an even number of Zings, A would be in group #1, B in group #2, C in group #1 again, and so on. At the end of the circle, A would have an unused x enmity and 'A-1', in group #2, would have an unused y enmity. Co-joining these would give each Zing, at most, one z enmity within the group.

Now assume an odd number of zings. A and 'A-1' would be in the same group, in the same way that a numbered circle with an odd number of members must have two odd numbers adjacent. (A-1 could not be in the other group from A as it already shares an enmity with A-2) But if A-1 co-joins enmities with A - and A is the only one with an unused x enmity for A-1's y enmity to co-join with - it can not be guaranteed that A, or A-1, dos not also have a z enmity within that group.

The long walk home:

Quote, "The man starts to walk at 14:00. Since the walk saved 10 minutes in the end, those 10 would be spent driving otherwise, meaning 5 minutes in each direction. So when the man met with his wife, it was 14:55 (saved 5 minutes before 15:00). He was walking for 55 minutes."

Just a minute. blow the whistle, call for a timeout and send for a moderator. Following advice from Iacomus, I am not saying you are wrong. Nevertheless, are you right

Now, whatever the length of journey, his wife, as you say has to save five minutes each way. As she would set off at the usual time, she would have to meet him five minutes drive from the station.

Therefore the normal journey time, has to be 10 minutes drive each way. She would then have to set off at 2.50pm (to be at the station by 3pm)
If she met him five minutes later, he would have walked for 1.05 hours.
Can you assist?

I only ask this question, in the sprit of fostering international relations with folks in Europe, you understand.

Quote, "You are on the ball. The 'odd street' method breaks down when loops are involved. I missed that."
Sorry, to be so slow. But where is the 'odd street' method mentioned?

ARGH!

I've just spent half an hour explaining the Zings and Iacomus' objection, and have meanwhile disconnected from Internet.

When I finished the monumental, 4-volume post, I then clicked 'Post' and guess what - Post.php cannot find server eerror microsoft explorer etc.
I reconnected but to no avail, the post got lost! It's the damn edit box and it loses contents if can't be posted!
AAARGH!!!

Anyway, I admit I've found an issue with my original solution but have resolved it - it's just a little addition.

Iacomus : I will re-work this some other time, but please see 2b2 section of the solution. Remember this is induction, so yoiu must treat A as the Zing that is being added to otherwise grouped party.

Try

From 'Missouri Mule' to 'Philadelphia Shyster' in one easy move!)

If it is dubious in some way to call the 'streets walked an uneven number of times' 'odd streets', which I agree does have an air of suggesting the streets are in some way peculiar, why then were you silent when they were called 'unevenly walked streets', which questions the sobriety of the walker/s, or the 'oddly walked streets', about which I make no comment other than 'if consenting adults choose to walk that way . . .'? Just what do you have against non-conformist streets anyway, and why? What harm did they ever do to you?

Relative

If the only problem is whether it is 'A' or 'A-1' that is being added to the otherwise-completed sorting, that is easily adjusted. I just re-label my Zings such that A becomes, B, B becomes C, etc etc, and A-1 adopts the label A, which has become vacant.

However, I can see no reason why I 'have to' follow your methodology in order to show that your method does not infallibly lead to a true conclusion. It is enough, and always has been enough, to provide a counter-argument that says in effect, 'there is one case where your argument cannot be guaranteed to hold.'