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?!?

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?

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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?

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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.'