The worlds first riddle!

Mark:
Way off!!! The answer is approximately 2.5*10^158
Look at the first few terms:
100 + 100*99 + 100*99*98
These total 980,200, they're only 3% of the terms, AND they're growing. The last term is 100!

Yes, you are correct. As usual.
I must read the question more carefully in future.

A swimmer 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

CANAL
1 km/hr

PALINDROMIAL

What is p(x)?


Every solution is of the form a(x + 1)n, for a any constant, and n any positive integer.
Let p(x) = an xn+ an -1 xn - 1+ ... + a2 x2+ a1 x + a0 be a palindromial whose derivative p'(x) = nan xn - 1+ (n - 1)an - 1 xn - 2+ ... + 2a2 x + a1 is also a palindromial. From p(x) we get an= a0 and from p'(x) we get nan= a1.
Now an -1 is determined by p(x) and the value of a1, from which in turn a2 is determined from p'(x). Continuing in this fashion, we see that all coefficients are determined by the value of an.

Since multiplying a palindromial by a constant doesn't change its citizenship as a palindromial, with the same true of its derivative, we conclude that there is a unique palindromial with leading coefficient 1 whose derivative is also a palindromial. A moment's thought will verify that the polynomials (x + 1)n fit the requirements.


Mark:
CANAL
1 km/hr

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.



Suppose a game of volleyball is tied at a score of 19 / 19. What is the probability that the game ends with a score of 21 / 19

Some particulars:
To win a game of volleyball a team must score 21 points and win by a margin of at least 2 points.

A team winning a rally either (a) scores a point if it served the ball, or (b) gains the right to serve the next rally if it received the ball.

In order to win a point a team must be serving the ball.




What is the smallest positive integer x so that 2168 divides x2003 + 1

PALINDROMIAL
Try wrote: "A moment's thought will verify that the polynomials (x + 1)n fit the requirements."

I'm not sure what this means. Do you agree with my answer that the coefficients of p(x) are found on the Nth line (where N is the order of p(x)) of Pascal's triangle?

In other words, if N is the order of p(x), then the coefficient of x^n is C(N, n) (N choose n).

VOLLEYBALL
Assuming each team has an even chance of winning each rally, the probability is 2/3.

One team (we don't care which or when) will go ahead 20-19 and will be serving. At that point, the probability of them (eventually) scoring again before the other team scores is 2/3.

2168 DIVIDES 2003X + 1

565

PALINDROMIAL

Mark:
"Do you agree with my answer that the coefficients of p(x) are found on the Nth line (where N is the order of p(x)) of Pascal's triangle?"

Nobody would disagree with that statement.

"In other words, if N is the order of p(x), then the coefficient of x^n is C(N, n) (N choose n). "

That will do nicely.

BTW I post any workings I have for information only, so that any passing reader may better understand the thinking behind the answer. I am sure by now everyone knows that should a ?'discrepancy' arise, it is a safer bet to accept your answer.


VOLLEYBALL
Assuming each team has an even chance of winning each rally, the probability is 2/3.




In the case of evenly matched teams, the probability is 2/3. This can be arrived at by use of a decision tree and summing of some infinite series, or by the following nifty analysis.

The only way for a score of 19-19 to result in a winning score of 21-19 is for the eventual winning team to reach a score of 20-19 and be serving the ball. Call this Event A. Let's call P the probability that the final score is 21-19, p the probability that the winning team wins any rally, and q = 1  p the probability that the losing team wins any rally. We must assume, of course, that p and q are constant throughout the play of the match.
Once Event A occurs there are two possible ways the match can proceed.

(a) The team in the lead and serving wins the next point, consequently ending the match. This occurs with probability p.

(b) The team in the lead and serving loses the next point, and consequently gives up the serve. It must then immediately win back the serve on the next rally, since otherwise the score would reach the forbidden 20-20. This puts us right back into event A again. This occurs with probability pq. Therefore P = p + pq P. Solving for P gives P = p/(1  pq). In the particular case that both teams are evenly matched, p = q = 1/2, and P = 2/3 as stated.


What is the smallest positive integer x so that 2168 divides x2003 + 1?
2168 DIVIDES 2003X + 1

565


I did not realise I posted this problem, as I was still working on it. However, so far I come up with:
First note that x2003 + 1 = (x + 1)(x2002 - x2001 + ... + x2 - x + 1). There are 2003 terms in the sum within the second set of parentheses, so this sum is odd. Therefore 2n divides x2003 + 1, then 2n must divide x + 1. Clearly, (or prehaps not) the smallest value for this to hold is x = 2n - 1.




Put a coin in a bottle and then stop the opening with a cork. How can you get the coin out of the bottle without pulling out the cork or breaking the bottle



Six schools each send their two top runners to a track meet. The twelve runners are assigned, at random, two runners to a lane on a six-lane track.

What are the chances that at least one school will have both of its runners assigned to the same lane




An ant is standing at one of the corners of a regular tetrahedron. Once each minute, the ant selects one of the adjacent corners at random and moves there.

What is the probability that the ant is back where it started one week after it starts its walk

2168 DIVIDES 2003X + 1
Oh! Was that supposed to be x^2003 + 1? It looked like multiplication.

BOTTLE
Push in the cork.

ANT
First thought: 1/4, but I need to work on this.

ANT
Yep - Close enough to 1/4 to be indistinguishable by most computers.

Mark:
2168 DIVIDES 2003X + 1
Oh! Was that supposed to be x^2003 + 1? It looked like multiplication.

You gave the right answer for the problem you saw. I have a problem in that all signs appear as ?'boxes' when posted. I will get it sorted.



BOTTLE
Push in the cork.

Damn, that was quick.


ANT
Yep - Close enough to 1/4 to be indistinguishable by most computers.



Each square on a 9 x 9 checkerboard contains a jumping beetle. At an agreed upon moment, they all jump from the square they are on to a diagonally adjacent square, that is, to a square which shares a single vertex with the square.

After they jump, what is the fewest number of checkerboard squares that must contain at least two jumping beetles




There are only two ?'T's in Timothy Tuttle.

True or false

True. Assuming your username is case-sensitive.

JUMPING BEETLES
I can get as low as 3.

TRACK TEAMS
I get 0.419 or 41.9%

DrewDad:
"True. Assuming your username is case-sensitive."

Clever


Mark:
JUMPING BEETLES
I can get as low as 3.

Color a 9x9 checkerboard so that the diagonals share the same color. Notice that since movement is diagonal, a beetle must jump to a square of the same color, thus allowing you to split the problem into smaller problems: the red beetles and the black beetles.
The red beetles can all be paired up, leaving no red squares with more than one beetle on it.

Now look at the black squares. Five rows of five squares alternate with four rows of four squares. Therefore, 25 beetles must be crammed into 16 spaces. Since at most 4 beetles can meet at a single square, the minimum number of squares where a beetle-meeting occurs is achieved when three squares have four beetles on them, and the rest have contain a single beetle, accounting for the total of 3•4 + 13 = 25 beetles.



TRACK TEAMS
I get 0.419 or 41.9%

Ditto





In how many ways can you change a
$50 bill using $20 bills, $10 bills,
$5 bills, $2 bills, and $1 bills




A man spent (in this order) a third of his life to date in the U.S., a sixth of it in India, 12 years in Egypt, half the remainder of his time in Australia, and as long in Canada as he spent in India.

Where did he spend his fortieth birthday




Dave drove at a steady clip along the highway, his wife beside him. "Have you noticed," he said, "that those annoying signs for Wild and Wonderful West Virginia seem to be regularly spaced along the road?
I wonder how far apart they are."

Sally glanced at her watch, then counted the number of signs they passed in one minute.

"What an odd coincidence!" exclaimed Sally "When you multiply that number by ten, it exactly equals the speed of your car in miles per hour."

Assuming that the car's speed is constant, that the signs are equally spaced, and that Sally's minute began and ended with the car midway between two signs, how far is it between one sign and the next

If I were him, I would like to spend it in a thebaid :wink:

40TH BIRTHDAY
In Egypt - he's 72

SIGNS
1/6 mile

Francis wrote, "If I were him, I would like to spend it in a the baid"

That reminds me of: A chicken and an egg are lying in bed. The chicken is smoking a cigarette with a satisfied smile on its face and the egg is frowning and looking a bit pissed off. The egg mutters, to no-one in particular, "Well, I guess we answered THAT question!"


40TH BIRTHDAY
In Egypt - he's 72

SIGNS
1/6 mile

Answer: The curious thing about this problem is that you do not need to know the car's speed to determine the spacing of the signs.

Let x be the number of signs passed in one minute. In an hour, the car will pass 60x signs. The speed of the car is 10x miles per hour. In 10x miles, it will pass 60x/10x, or 6 signs. The signs therefore are 1/6 mile apart.




This is guaranteed to drive you

Counting any way, except diagonally. How many times can you make the word: VALLEY



……........VALL
………....VALLE
……....VALLEY
……..VALLEY
…..VALLEY
..VALLEY
VALLEY

I'm not sure you got the meaning of "thebaid"

Francis, a thousand apologies. I thought it was a typo. I have enough trouble with English without ancient Egyptian. You caught me out with what we call an ?'Adrian'. A cryptic answer that is better than the question.

"The valley of the Nile, under Roman domination, was divided into four provinces: Lower and Upper Egypt, Lower and Upper Thebaid. The last two comprised the upper part of the valley. During the fourth to fifth centuries it was the chosen land of the monks, who by their sanctity and by the form they impressed on the monastic system greatly influenced the East and the West.

Their monasteries may be divided into two groups. The best known is the Pachomian group, founded and legislated for by St. Pachomius. They formed a real religious order with Tabenna as a mother-house and its superior as their general. Besides Tabenna there were Peboou, Schenesit, Akhmin, Esneh, Monchosis, Thebaid, Tesmine, Hermopolis, and Armoutim. Saint Pachomius governed this group till his death (346), and was succeeded by Abbot Orcisius, and then by Abbot Francis.

I should have known you can walk on water.

Between you and me let's not have an "Adrian" wall