The worlds first riddle!

There are trees hidden in each of the following sentences. How many wood you say

If I run, I'll catch up with my pal Derek.

There was a shop in every village we visited.

I feel my hair would look better if I let it grow another inch or two.

We raced around the park to a kiosk that sold crisps.

Answers to the questions I set.

Question 1.

He could not have taken part in any such election, so the answers is "No; he did not vote to join or not to join the EU".

The reasoning is based on two apparently insignificant pieces of information that were in the text; 'the family vault' and 'thin crescent moon'. A thin crescent moon can only be seen from the far north or far south of the Earth's surface, say Alaska or Northern Russia, and a Family Vault would only belong to a family that had a long history of living in that locality. No country from the far North has yet voted, or intends to vote in the foreseeable future, on joining the EU.


Question 2. As Try wrote in his reply, the pill-taker knows that he has two blue pills and one red pill in his hand. If he adds a second red pill and then halves each pill, keeping the two halves carefully separate, he now has two groups, each containing two red half-pills and two blue half pills, or one red and one blue, which is what was required.


Question 3

The question clearly says that the rabbit was imagined in the bottle, therefore it can be imagined out of the bottle

It has been brought to my attention that the standard of my questions has fallen of late. This is an assertion I am in no position to refute.

Therefore, the time has come, as the Walrus said, to say goodbye.
May I thank those that took part, and say, how much I enjoyed the experience.

I hope the last offering will go someway to redress the balance.



Two clocks were recently both wound up and set off together at 12 o'clock. They now both show 12.30.

One clock loses ten minutes ever hour and the other gains ten minutes every hour. What is the correct time


Mr. Brook's five daughters each gave books for Christmas to one or more of her sisters. Each presented four books and each received four books, but no two girls allocated her books in the same way.
That is, only one gave two books to one sister and two to another. Beth gave all her books to Alice; Christy gave three to Edith. Which sisters gave the four books to Deborah

An explorer wishes to cross a barren desert that requires 6 days to cross, but one man can only carry enough food for 4 days. What is the fewest number of other men required to help carry enough food for him to cross


How can three missionaries and three cannibals cross a river two at a time in a canoe. If the cannibals must never outnumber the missionaries left on one side, and two cannibals cannot paddle across together

Correct me if I am wrong but it seems to me that two of these problems - the clocks and the Cannibals - as given, do not have a solution.

The clocks.

If one loses as much time as the other gains, they can only tell the same time when one has lost a multiple of six hours and the other has gained the same, i.e. relative to ?'correct' time one has advanced half of the circle and the other has retreated half of the circle. If each gains/loses ten minutes per hour, they must gain/lose I hr every 6 hrs (10 mins per hour = 6 times 10 mins per 6 times I hr. Equals 60 mins per six hours equals 1 hr per six hrs) But before they coincide they must have each gained/lost six hours which requires ?'six times six' equals thirty-six hrs ?'correct' time.

So If we assume they start at 12;00 midday Sunday, both telling the same ?'correct' time, then they will again tell the same time at 12:00 Monday midnight, and both will show 6:00. The next time they coincide will be Wednesday midday and each will show the correct time.

So they can never, if they are started at 12:00 both showing the correct time, with one gaining and one losing 10 mins per hour, show the same time at 12:30.

It follows that if they do tell the same time at 12:30 after being started at 12:00 then they could not have been both set at the ?'correct' time at 12:00. No information was given as to what the alternative settings might be, and if we take ?'time x' as being the time they both showed at 12:30, then setting the clock that gains at ?'time x minus five minutes' and the clock that loses at ?'time x plus five minutes' then they will show the same time at 12:30 no matter what ?'time x' is chosen to be. So any answer at all is possible if they were not set at the ?'correct' time and we are not told what in fact they were set at.

It seems, therefore, that no unique answer to the clocks question exists.


The missionaries and the cannibals.

1. Cannibals must not outnumber missionaries on either bank
2. Two cannibals cannot be in the boat together.

We begin with 3 cannibals (C3) and 3 missionaries (M3) on the near bank. If one person takes the boat across he must then bring it back, and nothing has been gained, so two must go across and one return.

The boat holds two so this can only be 2M, 2C, or CM. But CC is not allowed (2. Above) and MM would leave M1 C3 on the near bank, which is not allowed (1. Above)

So the first trip has to be MC as all else is disallowed.

We now have C2 M2 on the near bank, M1 C1 and the boat on the far bank. The cannibal cannot bring the boat back as there would be C3 M2 on the near bank when he reached it.

So the return trip must be M1 as all else is disallowed.

We now have M3 C2 and the boat on the near bank, C1 on the far bank. Sending C2 in the boat is not allowed so again the next trip must be either M1 C1 or M2. If it were M2 then there would be M1 C2 on the near bank after the boat departed, and this is not allowed. If it were M1 C1 then there would be M1 C2 on the far bank when the boat arrived, and neither is this allowed.

So there cannot be a second trip across as all is disallowed.

One possible solution would be that a cannibal is tied and towed through the water, thus not being one of those in the boat. The difficulty is that if they could tie him to the back of the boat they could equally easily tie up all three cannibals on shore and then there is no problem to solve.


The desert.

One days travel equals one quarter of a pack.

Three men, A B and C, set off with full packs. At the end of day 1 they each have ¾ packs left. C gives A ¼ and B ¼, which fills both of their packs and leaves C with ¼ for the return trip.

A and B set out on day 2 with full packs. At the end of day 2 they each have ¾ packs. B gives A ¼ of his pack, filling A's pack and leaving B with ½ pack for the two days return trip.

A has 4 days left to travel and has a full pack with which to do it.

So it requires three men for the crossing to be made.


The books.

Let uppercase be the giver and lower case be the recipient.

A b1 c1 d1 e1
B a4
C d1 e3
D b2 c2
E b1 c1 d2

So each has received 4 books and each has given 4, but with no two givers having the same distribution.

D received one book from A, one book from C, and two books from E.

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

Try.

Maybe it was time for you to start a new thread anyway, somewhere on the ?'front cover' where you will catch more of the casual browsers.

CLOCKS:
As Iacomus has already found, the time is 12:30. The clocks can show the same time if and only if they were both set at 12:30.
The difference in distance travveled between the clock's pointers is 20 minutes per hour, or 1/3 of 1/12 of 360 degrees, that is 10 degrees per hour, or angle = 10*t.

The clocks show the same time when angle = n*360 and n is an integer. so we have
10*t = 360*n
t= 36*n , so from here follows that every 36 hours they are at the same position (3 days). This position is always the starting position.


BOOKS are an interesting puzzle. Only one possible solution: (in rows is who gives the books, in cols who receives them)

->>>edited because space not preserved
ABCDE
A 01 111
B 40 000
C 00 013
D 02 200
E 01 120

One additional (assumed) condition is needed : none of the girls can give any books to herself!
Now the possible book distributions are:
4 , 31 , 22 ,211,1111

Since A already has all four books (from B), she must be the one with 1111 distribution; any other girl would have to give a book to either herself or A to use 1111. OK, now E has all her books (1 from A, 3 from C). E must therefore use the 211 distribution, and D must use 22. Since D has only 1 book so far, the only way she can have four is to get 1 (still not given) from C, and the 2 from E. Now all is clear.


DESERT

ABC set off. They eat from A's supply 3 food on 1st day. C returns with 1 food left. AB eat from B's supply the next day. B goes back with 2 food, and A goes across with 4 food.

Why is this the minimum?
The man (C) must have full supply of 4 food 4 days before the end. Since he consumes 2 food by then, he must get 2 food from somebody else.
That somebody needs 2 food himself to return, so they have both 6 food on day 4.
They were at full capacity ,8 food, on day 3, so they alone could not make it and were given food by the third man necessarily . QED.


CANNIBALS:

As stated, there must not be more cannibals on the bank, not counting the boat itself, otherwise we have an impossibility (we must have an odd number of persons on one side at some point, and the rest on the other. One of them contains more C than M.)

So we have MMMCCC. MC goes across, leaving C there and returning. Picking up another C and leaving it on the other side.
We now have (* denotes boat which is excluded)
MMCC MC*
MMCC M* C
MMC MC* C
MMC M* CC

Now two M go across
MC MM* -> CC

and MC goes back

MC <- MC* MC

two M going back

CC MM*-> MC
CC M* MMC
then one M brings the other two C's.
C MC*-> MMC
MC*-> MMCC
MMMCCC

Thank you Tryagain, and I wish you all good in your future riddling Youn can always slip in another one when you wish so, and I will always have a go at it

Relative

You wrote:

"MC goes across, leaving C there and returning. Picking up another C and leaving it on the other side".

And when, in accordance with your post, the boat arrives at the far side for the second time there are two cannibals and one missionary on the far bank. This is not allowed.

The puzzle clearly states that at no time may the cannibals outnumber the missionaries. It says nothing of 'unless they have the boat in support'. :-)

I would query one point you made about the clocks. You say that after thirty-six hours the starting conditions would re-occur.

But if one clock is six hours ahead of the correct time and one is six hours behind they will tell the same time and it will not be the same as the starting conditions as each is six hours adrift of the starting conditions.

Once every thirty-six hours the clocks will tell the same time but this time will alternately be in keeping with the correct time and six hours adrift.

Iacomus : it is not indicated in the puzzle if the boat itself is considered to be on any particular side. The rules are about number of persons on the river bank. This is the only way that the transport is possible. Otherwise it is not. On this we agree.

Relative

The point I was making - and still make - is that it is not possible for the boat to ferry folks across the river without it at some time being on one shore or the other.

In your solution, from the moment the boat lands the second cannibal until it departs again there are two cannibals and one missionary on that shore. This has been expressly forbidden.

Iacomus: I understand your interpretation of the puzzle setting.

However, I don't agree with your statement:
"In your solution, from the moment the boat lands the second cannibal until it departs again there are two cannibals and one missionary on that shore. "

To 'be on shore' is a state, known to any sailor, as standing, two feet on the ground, ON the mentioned surface, and not floating in a canoe/boat in a river.

Relative

That sounds just a bit like a quibble to me. If a boat sails from London to New York, it is not considered to be still at sea just because not all of the crew and passengers are as yet on American soil.

But if you prefer I will withdraw the word 'ashore' and instead of ashore I will say 'at the far shore'.

It seems beyond reason to assume that the missionary is not outnumbered by the cannibals because he is sitting in the canoe against the bank, one of them is one metre away standing on the bank, and the other is transferring from one to the other.

If that were the case then the problem is easily solved; have all of the cannibals stand in the water up to their ankles. They would, by this reasoning, not be on any shore and therefore not capable of outnumbering anyone on either shore.

Or, following the same reasoning, let the missionaries stand in the water. The result is the same. Let everyone not in the boat stand in the water for that matter.

I don't have anything to add; under 'neutral boat' assumption there is a solution, without it there is no solution.

I wish no further debate whether this assumption is indeed included in the original problem setting, for this would be a task for a lawyer

Relative

Your first paragraph: we are in entire agreement.

Your second paragraph: Just as you wish.

Missionaries and Cannibals.

It is with some amusement that I note the replies from two great minds. However, as no one else has anything to add, may I offer a solution that may satisfy the majority?
Please note this answer assumes a person in the boat is not ?'on the bank'

Bank ?'A' Boat Bank ?'B'

MMM
CCC

MM >>>>>>>>M
CC >>>>>>>>C>>>>>>>>C

MM<<<<<<<<M
CC>>>>>>>>>>>>>>>>>>C

MM>>>>>>>>M
C>>>>>>>>>C>>>>>>>>>C

MM>>>>>>>>>>>>>>>>>>M
C<<<<<<<<<C>>>>>>>>>C

M>>>>>>>>>M>>>>>>>>>M
C>>>>>>>>>C>>>>>>>>>C

M>>>>>>>>>>>>>>>>>>>MM
C<<<<<<<<<C>>>>>>>>>C

>>>>>>>>>>M>>>>>>>>>MM
C>>>>>>>>>C>>>>>>>>>C

M<<<<<<<<<MM
C>>>>>>>>>>>>>>>>>>>CC

M>>>>>>>>>MM
C>>>>>>>>>CC

MMM
CCC


Books and Desert - correct.

Time:

The Answer is 12 o'clock.
In every hour the clocks differ by 20 minutes, in three hours by one hour.
The first time that they show the same time one must be 12 hours ahead of the other. They are therefore six hours adrift from the correct time, one ahead and the other behind. The correct time is 12 o'clock, 36 hours after they were set off.

Wood for the Trees: (Eight trees.)

1. iF I Run, I'll catch up with my pAL DERek
2. There wAS a sHoP IN Every village we visited.
3. I feEL My hair would look better if I let it grROW ANother inch or two.
4. We raCED ARound the park tO A Kiosk that sold crisps.


Light relief:

As we all know, a paradox is a self contradictory statement which conflicts the notion of what is possible.

The following statement is a well known paradox.

What happens when an irresistible force meets an immovable object?

If the force is irresistible then nothing can resist it. If the object is immovable, nothing can move it. Both entities cannot exist in the same universe so the question can't be answered.

One of the following is a paradox, the other is not.
(A) I always lie

(B) Is "No" the answer to the question?

Can you work out which one is the paradox, and why


A mathematician (R) and a philosopher (I), who went to college together, meet many years later. They have the following conversation:

R: "So did you get married and have kids?"

I: "Yes, I am married and have 3 kids."

R: "What are their ages?"

I: "Well" answered I slyly, "the product of their ages is 36 and the sum of their ages is the same as the number of the house where we played chess while at college!"

R: "I need more information", said R after thinking for a moment.

I: "The oldest looks a bit like you!"

Quickly, R tells I the ages of his three children.

Can you determine the ages of the three children All the information you need is contained above!


Riddle:
I was walking down Mulberry Lane
I met a man doing the same.
He tipped his hat and drew his cane,
and in this rhyme, I said his name.
What is the man's name

I bid you all a good weekend.

(A) I always lie

This is not a paradox, it's just a lie.

Mr (I) the philosopher has twin 2 year olds and a 9 year old.

The man in Mulberry lane is named Andrew.

I find it rather amusing that when posted with organ grinding monkeys, my solution to the river cross was deemed incorrect because I made the same assumption that you have now made Try. Namely that "a person in the boat is not 'on the bank'" Oh well, you win some and lose lots.

Try, I am glad you decided to post some more riddles.

I have a question concerning the 'Clock' riddle: In the riddle setup it is said that the two clocks


And the solution says it is now 12:00 -- I believe there is a typo in the riddle text "They now both show 12.30.". Am I correct?

The boat: Try's solution is clearly superior to the one I posted because stepping into/out of the boat is addressed clearly (my solution has problems with that )
Adrian: The boat problem traditionally has many faces. It was noted the first time you solved it that under the additional assumption of 'neutral boat' the solution is correct. However, the first riddle had a solution without this assumption, whereas the latest one doesn't.
The original problem I know for a long time goes like this:
A man, a goat, a wolf, cabbage, a boat. Only two can ride, and the man can row only. The wolf eats the goat, and the goat eats cabbage if left unattended on one side. How will they cross the river?

A) I always lie is a lie, as noted by Adrian. Meaning the person doesn't always lie, but does sometimes.
B) Is clearly a paradox. For this reason, the self-referential kind of language was banned from logic after the Russel paradox crisis ('Set of all sets').

" Oh well, you win some and lose lots."

Adrian, In this case, you win, win and win again. All three correct answers. I hope this goes some way to alleviate your sense of injustice. I am sure you Brits can bounce back from such an unfair result.

"Try, I am glad you decided to post some more riddles."

Relative, thank you for your kind sentiment. It is somewhat ironic that your post follows that of Adrian who with the comment, "Keep them coming" was responsible for the development of this thread.

Now, for an apology. I am sorry to say that I have destroyed the notes relating to the ?'Clock' puzzle.
"Two clocks were recently both wound up and set off together at 12 o'clock. They now both show 12.30.
One clock loses ten minutes ever hour and the other gains ten minutes every hour. What is the correct time?"

From memory, it is indeed 36 hours later. One would be 15 minutes fast, the other 15 minutes slow. Should I find the full correct answer, I will of course supply it.


Mathematician & Philosopher Solution
"The product of their ages is 36, and the sum of their ages is the same number as the house where we played chess."

The prime factors of 36 are 2 * 2 * 3 * 3 = 36, so either

(a) the three ages can be got from these four factors - for example,= 4 (=2*2), 3 and 3 is one possibility - or

(b) one of the children is aged 1, and the other two ages can be got from these factors, or

(c) two of the children are aged 1, and the other is aged 36 (unlikely).

This gives the following possibilities.
Product of Ages Sum of Ages
4 * 3 * 3 = 36 4 + 3 + 3 = 10
2 * 6 * 3 = 36 2 + 6 + 3 = 11
2 * 2 * 9 = 36 2 + 2 + 9 = 13
1 * 4 * 9 = 36 1 + 4 + 9 = 14
1 * 6 * 6 = 36 1 + 6 + 6 = 13
1 * 3 * 12 = 36 1 + 3 + 12 = 16
1 * 2 * 18 = 36 1 + 2 + 18 = 21
1 * 1 * 36 = 36 1 + 1 + 36 = 38
There are no other possibilities. Therefore, the house number must be 10, 11, 13, 14, 16, 21 or 38.

We know that this is not enough information for the Mathematician to solve the problem, even though he already knows the house number he is looking for. Therefore he must have found two possible solutions to the problem, and does not know which is correct.

House number 13 is the only one for which there is two solutions (2, 2, 9 and 1, 6, 6), and when the Philosopher refers to his oldest child we know that he cannot have a set of twins followed by a younger child, so the answer must be 2, 2 and 9.


One of the following is a paradox, the other is not.
(A) I always lie

(B) Is "No" the answer to the question?


The answer is B.
If you answer Yes that No is the correct answer then No is really the answer. If you answer No, then you imply that No is not the correct answer and that it should be Yes. This is impossible and the question is a paradox.

In example A, if this statement is true, then the statement is a lie and means that on least one occasion the person has told the truth. If the person is not telling the truth now, this also means he has told the truth on at least one occasion. Hence this statement is simply untrue and not a paradox.


In an effort to raise the standard from plain ?'difficult' to, ?'near impossible' I leave this for your consideration.

If I am spoken of as the first,
My peers are mostly greater than I.
Yet when I'm further down the order,
An excess of me means you could die.
Although I may rise and I may fall,
I have but one more claim to fame:
I am with the merchants all the time.
So I ask you now, what is my name


Two fathers and two sons went fishing. Each caught exactly one fish and yet there were only three fish caught. What is going on

Brown, Jones and Smith are a doctor, a lawyer, and a teacher. The teacher, who is an only child, earns the least money. Smith, who married Brown's sister, earns more than the lawyer. What is each man's job

Bad news, Try, travels fast.
But in case you are in doubt : I really thought you were going to quit!

I am usually terrible in word puzzles, but this one looks like bad luck.

Grandfather, father and son mame two fathers and two sons.
Jones teaches, Smith is a doc and Brown is a lawyer.

"I really thought you were going to quit!"

It is a simple fact that my supply of riddles and puzzles, is rapidly ending. Even Adrian complained that the math questions were of a grade two level. Therefore, excluding the more easy problems, the end is near. However, before that happens, I cannot possibly comment (at this time)on the validity of your answers.
Instead, I offer for your consideration the following.

Let us say you are a theoretical physicist who has just contacted an alien civilization using a brand new technology. Unfortunately, you cannot figure out quite yet where your new alien friends live. They could be on the other side of the Galaxy, across the Universe, or even in another dimension.

Nevertheless, you start talking, starting first in the language of mathematics, and then gradually working your way up to more complicated topics. After a while, you come to understand that their world is more or less like ours. They have the same concepts of math and geometry, the same physical laws, and their world seems in many ways like ours. It has the same chemicals, and orbits a star with numerous other planets. Moreover, the aliens are far more advanced than we are scientifically.

Eventually the aliens announce that they would like to transport you over to their world. Ignoring the details of how they plan on accomplishing this (hypothetically assume "magic"), they would like to make you feel at home when you arrive. So you starting telling the aliens everything you can about your home. Surprisingly, you discover that using only concepts built up from a shared understanding of math, physics, and chemistry, you can paint an amazingly detailed picture of life on Earth over time.

Assuming the aliens can build anything you can describe, and that you have the combined resources of all of Earth.

a) Could you theoretically describe your world well enough for you to survive on their world

b) Well enough, that you could not tell the difference

Try, I am sorry that I posted my answers here. I thought I sent a PM..