The worlds first riddle!

Thoh:

Soldiers
4 minutes.


Mark:

FIRST SOLDIER PROBLEM
L = length of column
Vd = dogs speed
Vm = men's speed

We want L/Vm.

L = (Vd - Vm) / 2
L = (Vd + Vm)

Vd/L - Vm/L = 1/2
Vd/L + Vm/L = 1

Therefore,
Vd/L = 3/4
Vm/L = 1/4

Since Vm/L = 1/4, L/Vm = 4 minutes.




Dog and a squirrel

This is just a shot in the dark; both the Dog & the Squirrel cover the same distance x during this time interval along the x-axis assumed to be along west direction. Then:

Integral from 0 to T of [V cos a dt]= U*T =X.........Eqn(1)

Where a is the instantaneous angle that dog's velocity makes with x-axis. Obviously a is dependent on the time variable t.


Speed of approach of the dog towards the squirrel is V - U cos a. In time T a distance D is covered with this velocity of approach. Hence...

Integral from 0 to T of [(V - U cos a)dt] = D ..........Eqn(2)
From Eqn(1) & (2) we have..
V*T - (U/V)*U T = D ...................................Eqn(3)

This gives T = D*V/(V^2 - U^2)

Bye, bye squirrel







Shari, a very rich lady goes with a bag of coins (less than 500) to the city of Phoenix. In the city she meets four mathematicians. She gives the first mathematician 4 coins and a quarter of the number of coins that are left in the bag. She applies the same procedure to the second, third and fourth mathematician.

How many coins did she leave home with

Try is correct about the dog. I got the same equation but i didn't use integrals. The time in which the dog catches the squirrel is the average of these 2 times: (1) the time in which dog catches the squirrel if they ran directly toward each other and (2) the time in which dog catches the squirrel if the squirrel ran directly away from the dog.

she had 244 coins at the start and 69 at the end

(69,244) is a solution to y=(256x+2100)/81

guess whose back. you can't get rid of me that easily.

BAG OF COINS
-12 (She may be rich, but she's in debt.)
Each mathematician gets 0 (4 + -4); so she's left with what she started with.

-268 (She's further in debt at the start.)
At the end she's got -93; so she comes out ahead!

HAHAHA i never thought of considering negative answers

Thoh:

she had 244 coins at the start and 69 at the end

(69,244) is a solution to y=(256x+2100)/81



Mark:

-12 (She may be rich, but she's in debt.)
Each mathematician gets 0 (4 + -4); so she's left with what she started with.

-268 (She's further in debt at the start.)
At the end she's got -93; so she comes out ahead!



Think positive, this is Shari we are talking about.





The lady went to the city and did indeed have 244 coins in her bag.

There are many ways to solve this problem. But there is at least one simple way to solve it. Suppose that the lady starts with 40 coins. The first vagabond gets 4 and 1/4 * 36 = 9, in total 13 coins. You would have found this number 13 immediately if the lady had brought 12 coins in addition (52) and then divided by four! This implies that if the lady brings 12 coins in addition, she had be able to divide 4 times by 4 without a remainder. So she should have brought 4 * 4 * 4 * 4 = 256 coins. But since we "borrowed" 12 coins, in reality the lady brought 256 - 12 = 244 coins to the city. (The other possibilities are 500, 756, ....)





Mark has become the owner of a piece of land, which to his dissatisfaction turned out to be nothing more than a big swamp. Mark wants to get rid of the swamp. A salesman advises him to use his fast-growing plants which can cover the swamp very quickly.

"This plant doubles every day, tomorrow you will have two, the day after tomorrow four, etc. In 80 days, your swamp will be completely covered."

Mark replies, "80 days? That takes far too much time, give me eight of these plants."



How many days did Mark save

[size=8]SWAMP
log2(8) = 3 days[/size]

Ok, this is trickier than a polecat in the hen house:

If the numbers 2^n and 5^n (where n is a positive integer) start with the same digit, what is this digit


Note; The numbers are written in decimal notation, with no leading zeroes.

[size=8]2^N AND 5^N
2^N * 5^N = 10^N

10^N has 1 as a leading digit; so for digit d to be a candidate, d has to be close to sqrt(1) or sqrt(10).

sqrt(1) = 1
Can d=1?
If the leading digit is a 1, then all trailing digits must be 0. This is not the case for either 2^N or 5^N.

sqrt(10) = 3.162+
Can d=3
Yes. In scientific notation, one number will have a mantissa between 0 and .162+, and the other number will have a mantissa between .162+ and .333...[/size]

wow... anyways... 2^5 = 32 and 5^5 = 3125 .....also 2^15 = 32768 and 5^15 = 30517578125 so mark is right

[size=8]HOME RUN
In the Nick of time, I would whisper in her ear that the proof includes the application of Vieta's formulas. (I cheated - that was way beyond me. But, hey, who wouldn't cheat for a Chick like that?)[/size]

HOME RUN

I did too!


A bag contains one red and one green disc. In a game a player pays $2 to play and takes a disc at random. If the disc is green then it is returned to the bag, one extra red disc is added, and another disc is taken. This game continues until a red disc is taken.

Once the game stops the player receives winnings equal in dollars to the number of red discs that were in the bag on that particular turn.

Who the heck wins and what is the exact value of the expected return on this game

[size=8]RED AND GREEN
Using Excel (instead of actually doing math), the expected return converges to e-1 = 1.718281828. As this is less than two, the game should not be played.

It's probably a series that sums to e, but is missing the first term (1) or something like that.[/size]

My riddle is...

What made Try disappear from Trivia?

[size=8]MISSILES
60000/60 = 1000 kilometers

UNITS
watt[/size]

Mark:

MISSILES
60000/60 = 1000 kilometers



One minute before they collide the missiles will be 1000 kilometers apart. The missile on the left is traveling at a speed of 38,000 kilometers per hour while the one on the right is going at 22,000 kph. Their relative speed is 60,000 kph. Since there are 60 minutes in an hour the speed is 1000 kilometers per minute, so one minute before they collide they must be 1000 kilometers apart.

The unnecessary information given was that the two missiles were 41,620 kilometers apart at the start, and where they would end up. All you need to do to solve the problem is to run the world backwards from the moment of the crash. At 1000 kilometers per minute, running in reverse will put the missiles 1000 km apart one minute before the collision.




UNITS
watt


watt is correct, it is a derived unit.




We all know that:

Continental drift speed is 1* 10^ -9 meter per second. This is equivalent to a speed of

(1)1 tm/s (2) 1 gm/s (3) 1 nm/s (4) 1 pm/s




What is always before you, yet you can never see it

[size=8]DRIFT SPEED
1 nm/s

BEFORE YOU
The future?[/size]