The worlds first riddle!

Hello! What do we have here?

Mark, "I think MOU meant 43. Since he left the last one for Francis, he ate only 21."

It would appear the MOU fan club are revolting.

MOU, "yeah Try, read more carefully"
Croatian translation; Wake up you SOB.

MOU:
Since I am huge fan of unpair numbers I would gladly leave the last one for Francis - so he will eat 22

we both eat 21, and one is left. I like unpair numbers, so I let Francis eat the last one.

Ok! Ok! But it is your fault, you let Francis eat the evidence.

Mark: ( Having calmed down the situation, by pouring gas onto troubled waters) Returns to his business like manner, and answered both questions before 9am. - Damn.


CHESS BOARD
204

BREAKDOWN
Allan. It takes Dillon as long to complete the last half of the trip as it takes Allan to complete the whole trip.



Can MOU swim?

MOU jumps from a bridge over a canal and swims 1 kilometer stream up. After that first kilometer, he passes a floating cork. He continues swimming for half an hour and then turns around and swims back to the bridge.

The swimmer and the cork arrive at the bridge at the same time. The swimmer has been swimming with constant speed.

The Question: How fast does the water in the canal flow
Does MOU get rescued?




What is going on here?

Whim lives in a street with house-numbers 8 up to and including 100. Paula wants to know at which number Whim lives.

She asks him: "Is your number larger than 50?"
Whim answers, but lies.

Upon this Paula asks: "Is your number a multiple of 4?"
Whim answers, but lies again.

Then Paula asks: "Is your number a square?"
Whim answers truthfully.

Upon this Paula says: "I know your number if you tell me whether the first digit is a 3."

Whim answers, but now we don't know whether he lies or speaks the truth.

Thereupon Paula says at which number she thinks Whim lives, but (of course) she is wrong.

The Question: What is Whims real house-number




Mark and a pirate are shipwrecked on an island with a boat which is large enough to carry only one of them to the mainland. They decide to let Lady Luck decide which one of them will be rescued.

On the island they find two coins. Assuming that at least one of the coins is fair -- that is, gives heads or tails 50% of the time -- but not knowing which one that may be, how can they use the coins to decide fairly who gets off the island

I dont eat that much

thank you
i'm fine
in the pediatric course we have no time to answer your riddles
just a time to breath


whim's house number is 81

Turn the book around - I think you're going backwards. :wink:

CORK and COINS sound recently familiar.

MOU is from town that lies on four rivers, so don't worry about me

Tyagain wrote:

She asks him: "Is your number larger than 50?"
Whim answers, but lies.

Upon this Paula asks: "Is your number a multiple of 4?"
Whim answers, but lies again.

The Question: What is Whims real house-number

I think the quwestion should be-

How much therapy is it going to take to get Whim to stop fabricating?

Here's a down payment for his therapist---->

I like helping people

Francis: "I don't eat that much"

You spend so much time in Lola's Café.


Usamashaker: "In the paediatric course we have no time to answer your riddles just a time to breath"

Keep going, you must almost be there. Remember, the longer you can keep them breathing, the longer they pay the bill.

"whim's house number is 81"




Note that Paula does not know that Whim sometimes lies. Paula reasons as if Whim speaks the truth. Because Paula says after her third question, that she knows his number if he tells her whether the first digit is a 3, we can conclude that after her first three questions, Paula still needs to choose between two numbers, one of which starts with a 3. A number that starts with a 3, must in this case be smaller than 50, so Whim's (lied) answer to Paula's first question was "No". Now there are four possibilities:


...........................................number is a square number is not a square
number is a multiple of 4............ 16, 36.............. 8, 12, 20, and more
number is not a multiple of 4...... 9, 25, 49.......... 10, 11, 13, and more

Only the combination "number is a multiple of 4" and "number is a square" results in two numbers, of which one starts with a 3. Whim's (lied) answer to Paula's second question therefore was "Yes", and Whim's (true) answer to Paula's third question was also "Yes".

In reality, Whim's number is larger than 50, not a multiple of 4, and a square. Of the squares larger than 50 and at most 100 (these are 64, 81, and 100), this only holds for 81.

Conclusion: Whim's real house-number is 81.


What made you smile today? Well, reading a reply from Bug's Bunny,
"I think the quwestion should be-"

Paula asks, "How much therapy is it going to take to get Whim to stop fabricating?"

It's not ?'fabricating that is the problem, it is ?'Fantasizing'. Your help is greatly appreciated, and will ensure a speedy recovery. :wink:


"MOU is from town that lies on four rivers, so don't worry about me."

It is obvious that the cork does not move relatively to the water (i.e. has the same speed as the water). So if the swimmer is swimming away from the cork for half an hour (up stream), it will take him another half hour to swim back to the cork again. Because the swimmer is swimming with constant speed (constant relatively to the speed of the water!) you can look at it as if the water in the river doesn't move, the cork doesn't move, and the swimmer swims a certain time away from the cork and then back. So in that one hour time, the cork has floated from 1 kilometer up stream to the bridge.

Conclusion: The water in the canal flows at a speed of 1 km/h.


STOP PRESS
MOU saved from very slow flowing river. Public holiday declared.



Mark:
Turn the book around - I think you're going backwards. CORK and COINS sound recently familiar. (So get a grip you SOB).

Ok. Message received and understood. Never seen before:



The playoffs in Major League Baseball begin in just about a week, culminating in the World Series which selects the champion team. The World Series is a tournament of seven games with the champion being the first team to win four games.

How many different ways can the World Series be played





Take a deck of 52 playing cards consecutively numbered from 1 to 52.
A perfect shuffle occurs when you divide the deck in two, one half in each hand, and riffle them together in an alternating fashion; after the shuffle the cards in the deck will be ordered as follows:
1, 27, 2, 28, 3, 29, ... , 24, 50, 25, 51, 26, 52

Notice that the top card always stays on top and the bottom card always stay on the bottom. Imagine now that you repeatedly shuffle the cards, performing a perfect shuffle every time. How many times will you have to do this before all the cards in the deck are restored to their original position

(You may want to start by convincing yourself that, indeed, the deck will return to its original position eventually.)

For the more adventurous: What happens if you put in the two joker cards and perform perfect shuffles with 54 cards; that is, how many times will you have to perform perfect shuffles with a deck of 54 cards before all the cards return to their original position

For the true daredevil: What happens with a deck of n cards

Tryagain wrote:

What made you smile today? Well, reading a reply from Bug's Bunny,
"I think the quwestion should be-"


-----------------------------------------------------------------------------------
Tragain wrote:

How many times will you have to do this before all the cards in the deck are restored to their original position

<paulaj thinks real hard,...almost hurts self > Answer-

A bagillion times?

This is the last time I'm going to derail this thread... [size=7]today.[/size]

8



52

really not sure what you had in mind with World Series question

World Series can have 8 possible outcomes (4:0, 4:1, 4:2, 4:3, 3:4, 2:4, 1:4, 0:4) if that's what you had in mind

And, if I take your question literally then it can be played in four ways - 4, 5, 6, or 7 games (that are, no matter the score, played on previously decided stadiums /decided by regular season score/).

I would never refer to you as an SOB!

WORLD SERIES
I get 70:
1 way to sweep
4 ways to win 4-1
10 ways to win 4-2
20 ways to win 4-3

Double the 35 (1+4+10+20) to get 70 because either team can be victorious.

CARDS (n)
If n=2^m, then m shuffles are required.
If n=2^m + 2, then 2m shuffles seem to be required.
Also, the number of shuffles is at most n-2.
Other than that, I don't know.

what i noticed is

the cards move with a pattern

in the first half of the cards
the next position = the current position*2 -1

in the second half
the next position = (the current position-n/2)*2

so each card moves a number of times to reach it's original location

if we have 4 cards

the fiirst card positions will be
1.....1......1.......1


the second card positios
2....3.....2

the 3rd card positions
3....2.....3

the 4th
4...4....4

so with 2 shakes the original arrangement will return

always the 2nd card takes the same number
of shakes that we need to return all cards to the original locatio

so with 52 cards

the 2nd cards positions
2
3
5
9
17
33
14
27
2


8 shakes

the same with any even number of cards

well, yeah, that's true also

Apologies for my late appearance. I could not log on with Explorer, I had to load up Firefox. However my password is saved in Ex. So I still could not login. A bug in the works methinks.



CARDS:
<paulaj thinks real hard,...almost hurts self >

?'Don't worry, we have a Doctor in the house.' :wink:

Answer- A bagillion times?... ?'Oh Paula sooooo close.'

This is the last time I'm going to derail this thread... today…
' You will have to go some, it has been heading for the bumpers for a long time.'


Usamashaker: (Good answer)

Cards

8
52


After a little experimentation, you'll see that the card originally in the i 'th position will have moved to the position given by the function p(i) defined by

p(i) = { 2i - 1, for i between 1 and 26
2(i-26), for i between 27 and 52


You can now follow along and see, for example, that card 1 returns to the first spot, while card 2 migrates in order through positions 2 - 3 - 5 - 9 - 17 - 33 - 14 - 27 and then returns to position 2. The following table lists the movement of the cards.

1 - 1
2 - 3 - 5 - 9 - 17 - 33 - 14 - 27 - 2
4 - 7 - 13 - 25 - 49 - 46 - 40 - 28 - 4
6 - 11 - 21 - 41 - 30 - 8 - 15 - 29 - 6
10 - 19 - 37 - 22 - 43 - 34 - 16 - 31- 10
12 - 23 - 45 - 38 - 24 - 47 - 42 - 32 - 12
20 - 39 - 26 - 51 - 50 - 48 - 44 - 36 - 20
18 - 35 - 18
52 - 52

You can immediately conclude that the deck will return to its original state after only 8 shuffles. Among the many curious patterns, notice how cards 18 and 35 flip positions after each shuffle.

The situation is much different for 54 cards; a similar analysis gives that
you'll need 52 shuffles in order to return the deck to its initial state. Moreover, each card (other than the top or bottom card) will migrate through each of the other 52 positions of the deck!

For the general case, it becomes useful to renumber the deck of n cards from 0 to n-1. The equation for the shuffle then becomes

p(i) = 2i (modulo n-1)
This gives that the number of shuffles is the order of the residue 2 in the group of units, under multiplication, modulo n-1, from which you can derive the following table for the number of shuffles.

n ………4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50

number of shuffles …
2 4 3. 6 10 12. 4.. 8 18.. 6 11 20 18 28 5 10 12 36 12 20 14 12 23 21




MOU
really not sure what you had in mind with World Series question

World Series can have 8 possible outcomes (4:0, 4:1, 4:2, 4:3, 3:4, 2:4, 1:4, 0:4) if that's what you had in mind. ?'That was the cards'

And, if I take your question literally then it can be played in four ways - 4, 5, 6, or 7 games (that are, no matter the score, played on previously decided stadiums /decided by regular season score/).


Not the way I was thinking.


Mark:
I would never refer to you as an SOB!

Sorry buddy, I have been watching too many John Wayne movies.

Mark:

WORLD SERIES
I get 70:

1 way to sweep
4 ways to win 4-1
10 ways to win 4-2
20 ways to win 4-3

Double the 35 (1+4+10+20) to get 70 because either team can be victorious.


Yes! He is right.

In order to count how many ways one team could win the World Series you only have to count how many ways that team could win
four out of seven games. This is computed by the binomial coefficient 7 choose 4, which has the value


7!
-----= 35
4!3!

Since there are two teams, you have to double this number to give you 70 ways for the World Series to be played.




This may be a duplicate:
The captain of a luxury cruise ship wants to install windshield wipers on the portholes of the ship. The straight wipers are to be attached at a point on the circumference of the circular portholes, furthermore, the entire length of the blade must remain in contact with the flat glass at all times.

How long should the wiper blade be so as to clean half of the porthole



Here's a curiosity:
(30 + 25)2 = 3025.

What other occurrences of this are there


Are there any such occurrences with numbers of more than two digits




Put all the positive integers into a basket. Reach into the basket and pull one out. Write down the number and return it to the basket. Repeat this process until you have pulled out a total of 1998 numbers.

Must some subset of the numbers you have selected have a sum divisible by 1998

CURIOSITY
20 and 30 are the only numbers that work.

BASKET
I'm fairly certain that the answer is yes for any positive N. I'm not sure how to prove it, but it is easy to show for small N. The problem is the same if you limit the numbers in the basket to 1 through N.

PORTHOLE
There are two choices:
sqrt(2)r
sqrt(3)r
where r is the radius of the porthole.

Mark I think my answers are on the same track, there are a number of ways to look at the problem.



Porthole

There are numerous variations on the main equation to the problem. If we assume (as we may) that the radius of the porthole is 1 unit, then the area, A, swept out by a wiper of length L is given by the equation
A = L2arccos(L/2).
Solving for A to be half the area of the porthole gives two solutions:
L = 21/2, 31/2.
A little straightforward calculus gives a maximum of about 52.45% of the area of the porthole with a wiper of length approximately L = 1.588.



The only other solution(s) with two digit numbers are (20 + 25)2 = 2025, and a generously interpreted (98 + 01)2 = 9801.

To arrive at the answer one can simply check all possible pairs of two digit numbers -- a daunting task by hand, but a simple task when armed with just about any programmable calculator. Alternatively, one can proceed as follow.

We want to solve the equation

(a + b)2 = 100a + b,
where a and b are two digit numbers. Reading the equation modulo 9 and modulo 11 gives the two equations
(a + b)2 = a + b (mod 9) and (a + b)2 = a + b (mod 11),
These equations give you that a + b has remainder 0 or 1 upon division by 9, and 0 or 1 upon division by 11. This restricts a + b to be one of 45, 55, or 99; note that (a + b)2 is a four digit number so a + b is at most 100. Squaring each of these gives the solutions above.

Solutions with integers of more than two digits can be handled in either of the ways above -- brute force, or a little number theory.




A sampling of some of the more interesting oddities:


(4 + 3 + 1 + 649)2 = 431649
(842 + 72 + 4)2 = 842724
(585886298 + 179545801)2 = 585886298179545801
a = 25*10n-2 +,- 5*10(n/2)-1, b = 25*10n-2, n even
(9 + 11 + 25)3 = 91125

(17147 + 18793 + 19616)3 = 171471879319616.




Integers

Let A1,...,A1998 be the numbers selected, and let Sk = A1 + ... + Ak. If any one of the Sk is divisible by 1998, we're done. In the contrary case, all of the Sk have a non-zero remainder when divided by 1998.
By the pigeonhole principle (there are 1998 sums but only 1997 possible remainder -- 1 to 1997) two of the sums must have the same remainder upon division by 1998. The difference of these two sums is divisible by 1998.





Recently I went on a vacation with my wife and her extended family. The dining room staff at the resort said that the 14 of us could not be seated at the same table, so they put us at two tables, of 8 and 6 seats respectively. So I wondered, if everyone wants to sit at least once with everyone else, how many meals do we need to accomplish this

When I say "sit with," I don't mean "sit next to", I just mean sit at the same table.

This leads to a more general question. Given any two numbers M and N, if M+N people are seated at tables of M and N respectively, how many meals does it take

This is not a trick question. At a given meal you have to sit in one seat throughout the meal, the tables do not overlap, etc.)




One night, in a month of the spring season, a certain young lady was lovingly happy with her husband in a big mansion, white as the moon, set in a pleasure garden with trees bent down with flowers and fruits, and resonant with the sweet sounds of parrots, cuckoos and bees which were all intoxicated with the honey of the flowers.

Then, on a love- quarrel arising between husband and wife, her pearl necklace was broken. One third of the pearls were collected by the maid-servant, one sixth fell on the bed - then half of what remained and half of what remained thereafter and again one half of what remained thereafter and so on, six times in all, fell scattered everywhere. 1,161 pearls were still left on the string; how many pearls had there been in the necklace




At 3:00pm, a boat is 12.5 miles due west of a radar station and traveling at 11mph in a direction that is 57.3 degrees south of an east-west line. At what time will the boat be closest to the radar station





Solve (if anyone can, which I doubt) the ff. systems:

a) x+y+2z=-1, 2xy+3z=0, -y+z=2
b) x+y+2z=1, 3x-y+z=-1, -x+3y+4z=-1
c) x-y+z-2w=3, -x+y+z+w=2, -x+2y+2z-w=9, x-y+2z+w=2

NECKLACE
If I kept up with the problem correctly, there were 128*1161=148608.

BOAT
3:36:50.07pm