The worlds first riddle!

Rap:

Working for the Man

Tp is pleasure time
Tw is work Time
I=R*Tw
M=I*Tp=R*Tw*Tp
Tp+Tw=80
Take Differentials of both
dM=0=R(Tp*dTw+Tw*dTp)
Tp*dTw=-Tw*dTp
dTw/dTp=-Tw/Tp
dTp+dTw=0
dTw=-dTp
dTw/dTp=-1
-Tw/Tp=-1
Tw=Tp
Since Tw+Tp=80 & Tw=Tp then Tw=40
So maximum is to work 40 hrs
Check Tw=40hrs, Tp=40 hrs M=R*40*40=1600R
Tw=41 hrs, Tp=39 hrs, M=R*39*41=1599R
It's a max Ans 40 hrs


40 hours. Isn't this an interesting coincidence, or is it?

Le U be your function the how much you get to enjoy your money per week.

Let h be the number of hours worked per week.

U = h*(80-h).
U' = 80-2h.
Set the derivative equal to 0:
80-2h=0, thus h=40.

Seems not!


Elsewhere it was written, "If I did okay, how would a brainy person like Rap do..."

Congratulations dude; fame at last!


Mark:

CARDS
I see what I did wrong.

Dang; you had me convinced you were right! :wink:



How many sub squares can you form on chessboard with n*n unit squares

NOTE: Sub squares must be an integer squared.

Hi Try,,, when are you going to venture up here to haverhill? I miss you ~

The computer i use is working funky like. There is a new one on the way.

So how are you?

[size=7]a 1x1 board has no subsquares, so n is at least 2
2x2 has 4 (1x1) = 4 total
3x3 has 9 (1x1) +4 (2x2) = 13 total
4x4 has 16 (1x1)+9 (2x2) +4(3x3) = 29 total etc
n / squares
2 4
3 13
4 29
an^3 + bn^2 + cn + d = total squares
i used a matrix to solve and got
total subsquares = (n^3)/3 + (n^2)/2 + n/6 - 1 [/size]

Someone etc: "40 hours. Isn't this an interesting coincidence, or is it?" What they said.

When i said i like being poor,,, i was trying to teach myself to accept my poorness,,, i used to think someone put a curse on me,,, ya know,,, made me not fall in love with rich/wealthy men in real life.

I wish that person would stop that if it's good,,, ~ forever Amen.

Thoh:

Sub Squares

a 1x1 board has no subsquares, so n is at least 2
2x2 has 4 (1x1) = 4 total
3x3 has 9 (1x1) +4 (2x2) = 13 total
4x4 has 16 (1x1)+9 (2x2) +4(3x3) = 29 total etc
n / squares
2 4
3 13
4 29
an^3 + bn^2 + cn + d = total squares
i used a matrix to solve and got
total subsquares = (n^3)/3 + (n^2)/2 + n/6 - 1

Would you say we were pretty close!

You can form n^2 unit squares.
You can form (n-1)^2 2 by 2 squares.
You can form (n-2)^2 3 by 3 squares.
You can form (n-3)^2 4 by 4 squares.
. . . You can form 1^2 n by n square.
The sum of all these sums is n*(n+1)*(2n+1)/6


Jovay, I hope you get your new computer. I will let you know if I am passing your neck of the woods!

Remember what Jebus said, "It is easier for a camel to pass through the eye of a needle than for a rich man to ascend to heaven." I expect he meant woman too!

I am well, but what about yourself?

Is it possible to show that any prime number other than 2 can be expressed as the difference of two squares, where each square is an integer squared

DIFFERENCE OF SQUARES
[size=7]Absolutely. Squares are separated by consecutive odd numbers:
Squares:
0, 1, 4, 9, 16, 25, 36, ...
Differences:
1, 3, 5, 7, 9, 11, ...
Therefore, all odd numbers can be expressed as the difference of two (adjacent) squares. Since all primes, except 2, are odd, they can be expressed as the difference of two squares.

For the above to be a proof, I'd need to show that any odd number is the difference of two squares.

Let M = 2N+1 for any integer N

(N+1)^2 - N^2 = (N^2 + 2N + 1) - N^2 = 2N+1 = M

QED[/size]

Mark:

DIFFERENCE OF SQUARES

Absolutely. Squares are separated by consecutive odd numbers:
Squares:
0, 1, 4, 9, 16, 25, 36, ...

Differences:
1, 3, 5, 7, 9, 11, ...

Therefore, all odd numbers can be expressed as the difference of two (adjacent) squares. Since all primes, except 2, are odd, they can be expressed as the difference of two squares.

For the above to be a proof, I'd need to show that any odd number is the difference of two squares.

Let M = 2N+1 for any integer N
(N+1)^2 - N^2 = (N^2 + 2N + 1) - N^2 = 2N+1 = M

QED

Let me check:

Let p be the prime number.
Let a=(p+1)/2, b=(p-1)/2.
Unless p is 2, p is the difference between a^2 and b^2:
A^2 - b^2 = (p^2+2p+1)/4 - (p^2-2p+1)/4 = 4p/4 = p.

Yep! It works!



If it takes Rap 20 hours to paint a house and it takes Mark 30 hours; how long will it take if they work together yet independently

[size=7]12 hrs[/size]

Rap

Paint It Black with Sympathy For The Devil to Gimme Shelter

The reason rap paints X 20ths of the house in the time it takes markr to paint X 30ths is that rap is already dressed for success.

x/20 + x/30 = bingo or housie housie

Try wrote: "I am well, but what about yourself?"

Hi Try,,,it's me pj. I have a compromised memory and keep forgetting my password etc.-etc.,,, so i had to log in with a new name. :wink:

I better than some, and worse than others thank you.

I am getting used to this new computer,,, i like it,,, check out this song if you have the time ~

"White Zombie - Electric Head Pt. 1 (The Agony) Music Video" It's church music to me ears :wink:

(i love this song! <rolls>

Rap:
Paint a house

12 hrs

Slippy:

The reason rap paints X 20ths of the house in the time it takes markr to paint X 30ths is that rap is already dressed for success.

x/20 + x/30 = bingo or housie housie


True!

Rap can paint at a rate of 1/20th of a house per hour.
Mark can paint at a rate of 1/30th of a house per hour.
Combine their efforts and they can paint 1/20 + 1/30 = 1/12th of a house in an hour.

So if their rate is 1/12th of a house per hour it will take 12 hours to paint an entire house.


Hi Jove, nice name! :wink:

Thanks for the song; I was listening to - Peter and Gordon ~ Please lock me away!

Good luck with the new computer.



A phial begins with one amoeba. Every minute, every amoeba turns into 0, 1, 2, or 3 amoebae with probability 25% for each case (dies, does nothing, splits into 2, or splits into 3).

What is the probability that the amoeba population eventually dies out

AHEM.....

"Oh Stormy...are you still on this thread..? ."
"Oh Izzie... I don't think so..."
"Oh Stormy...why is that...?"
"Oh Izzie...there are no letters for us girls..."
"Oh Stormy..let's just have tea then :wink: "

LGTIVTRGS

AMOEBA
[size=7]I don't know, but it is greater than 0.36972 because that's the probability that they will all die within 3 minutes.[/size]

Mark:

AMOEBA
I don't know, but it is greater than 0.36972 because that's the probability that they will all die within 3 minutes.

Even when he doesn't know, he is better than the rest of the world!


So what is the freekin' answer I hear ya'll shout…

Let p be the probability that just one amoeba eventually dies out, and all its descendants. The probability of n amoeba all dying out is p^n.

After the first turn there are four possibilities for the number of amoeba left, 0, 1, 2, 3 with probabilities of eventually dying out of 0, p, p^2, and p^3. With each outcome being equally likely the probability of all amoeba eventually dying out is 1/4*[ 1 + p + p^2 + p^3 ].

So p = 1/4*[ 1 + p + p^2 + p^3 ] or 1 -3p + p^2 + p^3 = 0.
This reduces to (p - 1)(p^2 + 2p - 1) = 0.

The solutions for p are 1, (sqr(2)-1), and (-sqr(2)-1). The only one which satisfies the constraints of the problem is sqr(2)-1 =~ .414213562

Which is indeed more than three!



Lzzie wrote, "AHEM....."


Nuff said…


ESBUSS

THEUMREASNT

BUSINESS

HUMAN INTEREST

Lzzie:

ESBUSS = BUSINESS

THEUMREASNT = HUMAN INTEREST

Well, that was 12 minutes of frantic fun…11 minutes longer than usual!


Hey, there is a free gift in my breakfast cereal! The manufacturers say that the gift comes in four different colors, and encourage one to collect all four (thus eating lots of cereal).

Assuming there is an equal chance of getting any one of the colors, what is the expected number of boxes I must consume to get all four

Anyone think it's sixteen? 24? 36? More?

Pardon my protoplasm but in the amoeba question does anyone wonder how the invigilator can dismiss the p= 1 probability without any explanation other than 'constraints of the problem'?

JENTACULAR

The expected number of binomial trials is:

1/(4/4) +1/(3/4) + 1/(2/4) + 1/(1/4)

Guilty as charged

Take the p below

Slippy:

Breakfast:

The expected number of binomial trials is:

1/(4/4) +1/(3/4) + 1/(2/4) + 1/(1/4)




First; if the probability of an event happening is p then the mean number of trials to obtain a success is 1/p.




Since there must eventually be a success the sum of probabilities is:
p + pq + pq^2 + pq^3 + ... = 1.

The mean number of trials (m) is: m = p + 2pq + 3pq^2 + 4pq^3 + ...
mq = pq + 2pq^2 + 3pq^3 + ...
m-qm= p + pq + pq^2 + pq^3 + ...
m(1-q) = 1.
m = 1/(1-q) = 1/p.

Second, the answer to the problem can be express as the sum of the following:

• Number of trials to get a first toy
• Number of trials to get a second toy once you have one toy
• Number of trials to get a third toy once you have two toys
• Number of trials to get the final toy once you have three toys

The number of trials to get one toy is obviously one.

Once you have one the probability of getting a different toy in the next box is 3/4, thus the expected number of trials is 1/(3/4) = 4/3 to get the second toy.

By the same logic the number of trials to get the third is 1/(1/2)=2 and 1/(1/4)=4 for the final toy.

Summing these yields 1 + 4/3 + 2 + 4 = 25/3 = ~ 8.333 .


Slippy continues:


"Guilty as charged

Take the p below