The worlds first riddle!

Iacomus,

In my lost post, I analysed your example (a circle with odd number of Zings) once more although it is included in the solution as 2b2, to be more clear.
When you want to show a hole in a proof, you need to follow it exactly in the terminology of the proof, otherwise argument is out of context.

You do NOT need that if you want to just provide an example that is suspicious in a way that it seems the proof doesn't work for that case.

To check if this is in fact so, you need to run the proof on it, meaning you have to use the proof's terminology .

Anyway, once again, the conterexample is an odd circular path of enemies, which is handled in section 2b2 of the proof. I cannot see a problem with that.

I discovered an uncovered case in the orig. solution but it turns out it is solvable:
The 'third enemy' of a member of the circle can be himself a member of the same circle. Say we are considering circle member An, and his third enemy (we will use * to denote third enemy) An* = Am, some other member of the circle. m comes later in the circle than n. When we move An-1 to the same group with An, we 'offend' it - that is, An now has two enemies in the group: An-1 and An*. An+1 is in the other group. We move An, offending An+1, etc. until we come to Am. Am is NOT in the same group as An, when Am is offended, because we've already moved An away from Am. So Am-1 and Am are in the same group, while An = Am* and Am+1 are in the other group. The re-arrangement finishes here in this case.

When I have more time I will compile a clear solution to the Zing problem since it is very interesting and I agree there might be still holes in the solution.

Try:


I agree with that.



Hmm, I believe there is really no way of knowing the normal journey time. It could be anything - an hour, five minutes, ten minutes. In fact it is interesting that we can solve the puzzle without knowing the normal journey time.



Even if she did set off at 2.50pm she would have met him at 2.55 (normally at 3.00 but now 5 minutes earlier). So he would have walked for 55 minutes.

Now you owe me two beers. When are you coming to visit over here to improve Euro-American relations

Relative

re: Zings.

After some careful work I now have a proof; a rigorous and complete proof.

Unfortunately for my earlier claim it is a complete and rigorous proof that 'three-enmity Zings' can always be arranged into two groups such that no Zing has more then one enmity inside the group. (Oh well; that is the way proofs go sometimes! <<He includes whatever the emoticon is for a Gallic shrug>>)

The key piece that I had not factored in was this; - if each Zing has three enmities, and each enmity is shared mutually by two Zings (with 'Z' being the total number of Zings, and 'E' being the total number of enmities) then 3Z = 2E.

From this it follows that if 'Z' is an odd number then not all Zings in an odd number of Zings can have three enmities as this would require a fractional number of enmities. With at least one Zing in an odd number of Zings having two or fewer enmities, arranging an odd number of Zings into two groups such that no Zing has more than one enmity within his group can be done every time.

Iacomus, I will be most happy to see your proof!
I hope that you will post it here

Quote, "Even if she did set off at 2.50pm she would have met him at 2.55 (normally at 3.00 but now 5 minutes earlier). So he would have walked for 55 minutes."

It is not often I admit defeat in the face of adversity, and superior intellect. However, after discussing the matter with the woman concerned, she informs me that her journey time is in fact 1 hour 10 minutes each way. Therefore, this is not one of those times. :wink:

"Now you owe me two beers. When are you coming to visit over here to improve Euro-American relations"

As soon as you join the E.U. How well I remember Ljubljana, by the river on which the argonauts carried the Golden Fleece. A city of Renaissance, Baroque, and especially Art Nouveau facades. Those were the days.

Quote," From 'Missouri Mule' to 'Philadelphia Shyster' in one easy move!)"
Oh dear. What have I done now?

"Just what do you have against non-conformist streets anyway"

You may well ask. Have you noticed the way they change, from Streets to Roads, Drive, Avenues, Boulevards, Lanes etc. When is a street not a street?

Why I ask you, do you park on a drive, drive in a park, but never park when you drive?

"After some careful work I now have a proof; a rigorous and complete proof. "
"Unfortunately for my earlier claim… Oh well; that is the way proofs go sometimes!"

I think not.
Pure mathematics consists entirely of such asseverations as that, if such and such a proposition is true of anything, then such and such another proposition is true of that thing... It's essential not to discuss whether the proposition is really true, and not to mention what the anything is of which it is supposed to be true... If our hypothesis is about anything and not about some one or more particular things, then our deductions constitute mathematics. Thus mathematics may be defined as the subject in which we never know what we are talking about, nor whether what we are saying is true.

Although the ubiquity of people who neither know what they're talking about nor know whether what they're saying is true may incorrectly suggest that mathematical genius is rampant, the quote does give a succinct, albeit overstated, summary of the formal axiomatic approach to mathematics.

The Structure of a Proof
The basic structure of a proof is easy: it is just a series of statements, each one being either
• An assumption or
• A conclusion, clearly following from an assumption or previously proved result.
And that is all. Occasionally there will be the clarifying remark, but this is just for the reader and has no logical bearing on the structure of the proof.

A well written proof will flow. That is, the reader should feel as though they are being taken on a ride that takes them directly and inevitably to the desired conclusion without any distractions about irrelevant details. Each step should be clear or at least clearly justified. A good proof is easy to follow.

When you are finished with a proof, apply the above simple test to every sentence: is it clearly (a) an assumption or (b) a justified conclusion? If the sentence fails the test, maybe it doesn't belong in the proof.

Let's start with an example.

Example: Divisibility is Transitive
If a and b are two natural numbers, we say that a divides b if there is another natural number k such that b = a k. For example, 2917 divides 522143 because there is a natural number k (namely k = 179) such that 522143 = 2917 k.
Theorem. If a divides b and b divides c then a divides c.
Proof. By our assumptions, and the definition of divisibility, there are natural numbers k1 and k2 such that
b = a k1 and c = b k2.
Consequently,
c = b k2 = a k1 k2.
Let k = k1 k2. Now k is a natural number and c = a k, so by the definition of divisibility, a divides c.
q
If P, Then Q
Most theorems (homework or test problems) that you want to prove are either explicitly or implicity in the form "If P, Then Q". In the previous example, "P" was "If a divides b and b divides c" and "Q" was "a divides c". This is the standard form of a theorem (though it can be disguised). A direct poof should be thought of as a flow of implications beginning with "P" and ending with "Q".
P -> ... -> Q
Most proofs are (and should be) direct proofs. Always try direct proof first, unless you have a good reason not to.
It Seems Too Easy

If you find a simple proof, and you are convinced of its correctness, then don't be shy about. Many times proofs are simple and short.
In the theorem below, a perfect square is meant to be an integer in the form a2 where a itself is an integer and an odd integer is any integer in the form 2a+1 where a is an integer.
Theorem. Every odd integer is the difference of two perfect squares.
Proof. Suppose 2a+1 is an odd integer, then
2a+1 = (a+1)2 - a2.
q
Where's the proof? It's there. It's just very short.
Another Simple Example
Recall that a natural number is called composite if it is the product of other natural numbers all greater than 1. For example, the number 39481461 is composite since it is the product of 15489 and 2549.
Theorem. The number 100...01 (with 3n-1 zeros where n is an integer larger then 0) is composite.
Proof. We can rewrite our number as 100...01 = 103n + 1 where n is an integer larger than 0. Now use the identity a3 + b3 = (a+b)(a2 - a b + b2) with a = 10n and b = 1, to get
(10n)3 + 1 = (10n + 1)(102n - 10n + 1).
We will be done once we have shown that both factors (10n + 1) and (102n - 10n + 1) are greater than 1. In the first case, this is clear since 10n > 0 when n > 0. In the second case, 102n - 10n = 10n (10n - 1) > 0, when n > 0. This completes the proof.
q
Make sure you understand why it was neccessary to discuss the two cases at the end.
One-to-One Functions
A function f:X->Y is called one-to-one if for any pair a, b in X such that f(a) = f(b) then a = b. Also, if f:X->Y and g:Y->Z are two functions then the composition gf:X->Z is the function defined by gf(a) = g(f(a)) for every a in X. Note that the composition gf is only defined if the domain of f is contained in the range of g.
Theorem. If two one-to-one functions can be composed then their composition is one-to-one.
Proof. Let a and b be in X and assume gf(a) = gf(b). Thus, g(f(a)) = g(f(b)), and since g is one-to-one we may conclude that f(a) = f(b). Finally, since f is one-to-one, a = b.
q
Roots of Polynomials
A number r is called a root of the polynomial p(x) if p(r) = 0.
Theorem. If r1 and r2 are distinct roots of the polynomial p(x) = x2 + b x + c, then r1 + r2 = - b and r1 r2 = c.
Proof. It follows from our assumptions that p(x) will factor
p(x) = (x - r1) (x - r2)
If we expand the right hand side we get
p(x) = x2 - (r1 + r2) x + r1 r2.
Compare the coefficients above with those of p(x) = x2 + b x + c to get r1 + r2 = - b and r1 r2 = c.

I hope that may be some help to you in any future deliberations.

Relative :
"Now you owe me two beers. When are you coming to visit over here to improve Euro-American relations"
Try:
"As soon as you join the E.U."

Relative:
"GREAT!! So let me make a reservation for you in the hotel .. we are joining EU on 1st of May "

Looks like you've been here before .. any chance you've seen our Alps?

Now you make it soud that bad

There is little polution around here, you must be heavily outdated or misled.

Me and Frank don't agree on ice and freezers, but when he comes here, I'll buy him a beer or two.

a) yes, and so can I
b) 13

Answers

The 'transporting beers' question is superficially similar to Try's 'camel and bananas' question and was written to be that way in order to fog the issue a little. It is different in that the camel cannot eat all of the bananas it will need before it starts whereas R can allow T to take his payment before starting.

No matter how R arranges things he cannot guarantee receiving more than 1 000 beers so the simple course for him to follow is for him to tell T 'bring me 1 000 beers and keep the rest'. This has the curious result that his best option is to pay more than was asked and to pay it in advance of delivery.

If he follows the 'camel and bananas' paradigm he receives, approximately, just under 500 beers.


How much money did T have?
None at all.


"Strangely, the writing was not magnified".

This can be paraphrased as;
1) "The writing consisted of the word 'NOT' which was magnified"
2) [Adrian's answer] "The writing did not consist of the word 'magnified' but something else"

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Relative

The proof, as it stands, is somewhat long-winded for posting on A2K. However, I will slim it a little and then make it available to you.

Try

Thank you for the advice on writing a proof and, be assured, if I run into difficulties you will be the first I will call on.

Try

Predicting scores.

Yup; he'd be right every time.


Not containing an 'A';

Taking the question literally, around fifteen words, though I found that finding them and counting them conflicted a little. But if this is one of those 'word puzzles' then there are two words in 'fifteen seconds' that do not contain an 'A'; three if 'in' is included.

All of the Macs

The Scots have always been famed for engineering, so probably none of these are odd.

After a flying visit to Olde London town, I return with a clutch of English riddles:

Little Red Riding Hood was walking through the forest to take her granny some eggs. On the way, she had to cross a bridge with an angry troll guarding it. "Give me half your eggs and half an egg and I will let you pass," said the troll. This she did. By the time, she reached her granny's cottage she had to pass four bridges in total.

A troll had guarded each and each made the same demand, which Little Red Riding Hood met. By the time she reached her granny's cottage, she found she had no more eggs left. How many eggs did she start with
PS She did not have to split an egg.

One hundred bushels of corn were distributed among one hundred people, in such a way that each man received three bushels, each woman two bushels, and each child half a bushel.

Given that there are five times as many women as men, how many children are there

PM to Try re: eggs and bushels.



Question #1

John - we will call him John though we do not in fact know what his name was - John goes to put some flowers at the family vault as is his monthly custom. But this time he has been working overtime and it is well into the late evening when he sets out.

He heads out to the graveyard with his flowers, roses BTW. It is dark. The thin crescent moon and a sprinkling of stars are all the illumination he has, but he knows his way and is not concerned.

Unfortunately, at precisely 10:25 pm, he falls into a newly dug grave that was not there the last time he passed this way. He falls, hits his head forcibly on a shovel that has been left by the workmen ready to continue the next day, and dies.

Now here is the question - did he vote to join or not to join the EU?


Question #2

A man - we'll call him John though we do not in fact know what his name was - is trekking through the wilderness. This is most foolish of him as he is on medication. He has a blue bottle of pills and a red bottle of pills. Each day he must take one from the blue bottle and one from the red bottle. If he misses doing this he will be very ill and a wilderness is no place to be ill so he is very careful. If he takes more than one of either kind he will be more than very ill, he will be very dead.

So, one day he has tipped one pill from the red bottle and is in the process of tipping one from the blue bottle when he suddenly sneezes. To his horror he finds that as a result of the sneeze he must have tipped two from the blue bottle, as he now has three pills in his hand. So what is he to do? He cannot throw them away as he has to finish the complete course of treatment. The pills are identical in every way that he can check, and he does not know how many pills each bottle originally held.

What does he do?


Question #3

You have an extremely large bottle. Imagine a very young rabbit - we'll call him John though we do not in fact know what his name was -. in the bottle. The neck of the bottle is far too small for the rabbit to pass through, at least in one piece, and the bottle is no different from most other bottles in the number of necks and alternative means of getting in or out of the bottle. (It is no different from a typical glass bottle except for being a bit larger and having a rabbit in it)

Your task; to rescue the rabbit from the bottle without harming either the rabbit or the bottle. How do you do it?

PM three possible answers.

Try has requested that I post my replies to his questions, so here goes:

The eggs:

She began with 15 eggs.
After the first bridge she had 7 eggs
After the second bridge she had 3 eggs
After the third bridge she had 1 egg
After the fourth bridge she had 0 eggs


The bushels

There were 5 men (implies 15 bushels)
There were 25 women (implies 50 bushels)
There were 70 children (implies 35 bushels)

5 + 25 + 70 = 100 people
15 + 50 + 35 = 100 bushels


Lids, locks, and boxes.

Locks are locks of hair, lids are eyelids, and boxes are in the sense of 'a box around the ears'

A 3x3 puzzle consists of a list of nine words. These words can be grouped in groups of three so that each group shares a common associated word.

For example, Horse, Tennis, Lace: = Shoe.
The three words obtained like this can then be grouped to find a single word, which is the answer. E.g.

Africa
Deep
Dip
Face
Minnie
Oliver
Park
Puppy
Star

Face, Oliver, Star: = North
Dip, Minnie, Puppy: = Skinny
Africa, Deep, Park : = South

North, Skinny, South:= Dakota


Now try:

Away
Bath
Birth
Fin
Fish
Lemon
Nurse
Snake
Wing

3x3 answer:
Bath, Fish, Snake:= Oil
Away, Birth, Wing:= Right
Fin, Lemon, Nurse:= Shark

Oil, Right, Shark:= Whale

Well done Iacomus.


Riddle:

Was it a bar or a bat I saw?
Was it a car or a cat I saw?
Was it a hat I saw?
Was it a mat I saw?
Was it a rat I saw?
Was it a vat I saw?
Oh, who was it I saw? Oh, who?

Who am I

Palindpoem entitled

'Name not one man!'

Oh, who was it I saw? Oh, who?
Was it a "Potato Pat" I saw?
Noella? Cain? A maniacal Leon?
Mad Adam?
Dr Awkward?
"It is I?" I moan. "O Hannah; O Naomi; Is it I?

Can anyone produce a better poem/riddle in the same way?



What is the well known phrase or saying

a) If a large solid-hoofed mammal becomes available to you without compensation, refrain from casting your faculty for seeing into the oral cavity of such a creature.

b) Each vaporous mass suspended in the firmament has an interior decoration of metallic hue.

c) It is not advantageous to place the sum total of your barnyard collections into the same wicker receptacle.

d) Feathered bipeds of a kindred mind in their segregated environment associate with a high degree of amiability.

e) Deviation from the ordinary or common routine of existence is that which gives zest to man's cycle of existence.

f) He who locks himself into the arms of Morpheus promptly at eventide, and starts the day before it is officially announced by the rising sun, excels in physical fitness, increases his economic assets and celebrates with remarkable efficiency.