The worlds first riddle!

Markr's touching rejoinder was that the probability of at least one woman touching at least one man was 1 ( and a deliciously salacious codicil so it must have been a tapas bar).

contretemps

oo oo i got one:

solve for x

x+x/(x+x/(x+...=sqrt(x+sqrt(x+...

Thoh's oooh oooh
[size=7]On the right is a continued fraction
x+1/(x+x(X+.....=X+(1/x)*(1/x)*(1/x)....=x+(1/x)^infinity=x

on the left is an expansion
sqrt(x+sqrt(x+...=x^(1/2)+x^(1/4)+x^(1/8)+....
taking log of the left side
ln(x^(1/2)+x^(1/4)+x^(1/8)+....)=1/2*lnx+1/4*lnx+1/8lnx+...
1/2*lnx+1/4*lnx+1/8lnx+...=lnx(1/2+1/4+1/8+....)
lnx(1/2+1/4+1/8+....)=lnx(1/2*(1+1/2+1/4+1/8+...))
lnx(1/2*(1+1/2+1/4+1/8+...))=lnx(2*(1/(1-1/2))
lnx(2*(1/(1-1/2))=lnx(1)
antilog
exp(ln(x))=x

putting the right and the left together

x=x and 0=0

or maybe knot[/size]

Rap

Rap:

Students that tell the truth will answer that there favorite subject is math, science, or social studies only---students that lie will say that there favorite subject is any thing but their favorite subject and answer the other two. So no student will answer all three since that will be both a lie and the truth, and the students either lie or tell the truth not both. So if I were to draw three circles (sets) that represent the answers then (I'll call then A, B, & C) Then

A+B+C-A∩B-A∩C- B∩C+ A∩B∩C=200
A+B+C=250
So
A∩B+A∩C+ B∩C-A∩B∩C=50
But
A∩B∩C=0
Since these students lie and tell the truth
So
A∩B+A∩C+B∩C=50

Results 150 of the students tell the truth and 50 lie.


Mark:

STUDENTS
liars will answer yes twice; truthers will answer yes once

2L + T = 250 (number of responses)
L + T = 200 (number of students)
L = 50 (liars)
T = 150 (truthers)

Nice answers dudes!


Jovay, what a nice story. Reminds me of the feeding of thousands when Jebus asked; "How many loaves have ye, and how many packets of Cheezits? Go and see. And when they knew, they said, five, and two packets. - Ahem!


Hey Slippy!Thoh; wow! Where did that come from! It must be 0x0 or x0x or something close.

THOH
[size=7]Setting y=left side gives:
y^2 = x + y or x = y^2 - y

Setting y= right side gives:
y = x + x/y

Substituting x in the first for x in the second gives:
y = y^2 - y + (y^2 - y)/y or y = y^2 - 1

Solving for y gives:
y = (1+sqrt(5))/2 and (1-sqrt(5))/2

Solving for x gives:
x = 1 for both cases.

Therefore, x = 1.[/size]

Partial differential equations required.

Water flows into a tank at a rate of 1 gallon per second. Water leaves the tank at a rate of 1 gallon per second for each 100 gallons in the tank.

The tank is initially empty.

How long will it take for the tank to fill with 50 gallons of water

Tank filling problem

[size=7]Let V be the volume of the tank, let qin be the flow into the tank and qout be the flow out of the tank
From mass balance
qin-qout=chance in volume with respect to timwe (dV/dt)
qin-1g/s
qout=1g/s/100g*Vg
qin-qout=1g/s-1g/c/100g*Vg=(100-V)/100
(100-V)/100=dV/dt
separating variables
dV/(100-V)=dt/100
integrating
ln(100-V)=-t/100+C
boundaty condition @ t=0 V=0
C=ln(100)
Ln((100-V)/100)=-t/100
(100 -V)/100=exp(-t/100)
100-V=100exp(-t/100)
V=100(1-exp(-t/100))
Let V=50 solve for t
½=1-exp(-t/100)
exp(-t/100)=1/2
-t/100=ln(1/2)=-ln(2)
t=100*ln(2) sec = 69.3 sec[/size]

Rap

On the THOH problem, I reversed left and right. That doesn't change the answer, but it may have caused confusion when reading my solution.

Rap:

Tank filling problem

Let V be the volume of the tank, let qin be the flow into the tank and qout be the flow out of the tank
From mass balance
qin-qout=chance in volume with respect to timwe (dV/dt)
qin-1g/s
qout=1g/s/100g*Vg
qin-qout=1g/s-1g/c/100g*Vg=(100-V)/100
(100-V)/100=dV/dt
separating variables
dV/(100-V)=dt/100
integrating
ln(100-V)=-t/100+C
boundaty condition @ t=0 V=0
C=ln(100)
Ln((100-V)/100)=-t/100
(100 -V)/100=exp(-t/100)
100-V=100exp(-t/100)
V=100(1-exp(-t/100))
Let V=50 solve for t
½=1-exp(-t/100)
exp(-t/100)=1/2
-t/100=ln(1/2)=-ln(2)
t=100*ln(2) sec = 69.3 sec


Rock on Rap!

I have been asked for the answer to the light bulb problem (Page 1069). So just for the record; this is what I have:


Warning: This is just a quick explanation which assumes some knowledge of the exponential distribution and integral calculus.

The density function of the shorter lived life bulbs is f(x)=1/100 * e^-x/100.

The density function of the shorter lived life bulbs is f(x)=1/200 * e^-x/200. The survival is 1 minus the integral of the density function. S(x)=e^-x/100, and S(x)=e^-x/200.

The survival function for the entire group is (e^-x/100)^5 * (e^-x/200)^5.

The density function for the group is the derivative of 1 minus the survival function = e^-3x/40.

The mean for the group is the integral from 0 to infinity of x times the density function for the group = 40/3 hours




Probability and algebra required.

A drunk is on one point of an n-gon, such that n is an even number. The drunk moves along the perimeter of the n-gon. Each step takes him to an adjacent point, and every step is chosen at random.

What is the expected number of steps before he arrives at the point directly opposite that of his starting point (as a function of n)

DRUNK ON N-GON
I get (n/2)^2

To those that celebrate the supreme sacrifice; Happy Easter everyone. To those that don't, purgatory awaits - Amen



Mark:

DRUNK ON N-GON
I get (n/2)^2



Taking the ant and spider problem for the method to compute any given expected number. Use telescoping sums to find that the expected number at either point next to the final point is (n-1).

Use telescoping sums again to determine the expected number of the starting point as a function of the next to last point. I get: n^2/4 which some may say is not a whole lot different.



taenod


bobawckspl

taenod = anointed


bobawckspl = backspin bowl

To those that celebrate the supreme sacrifice; Happy Easter everyone. To those that don't, purgatory awaits - Amen

If you're lucky -

well that's a shorter explanation than what i had, but yes, 1 is the correct answer

Stormy:

taenod = anointed

bobawckspl = backspin bowl

Awesome!



What do you make of the sequence 1, 2, 3, 7, 43, 1807, 3263443,

[size=7]k
well
1 + 2 = 3
3 x 2 + 1 = 7
7 x 3 x 2 + 1 = 43
43 x 7 x 3 x 2 + 1 = 1807
1807 x 43 x 7 x 3 x 2 + 1 = 3263443
3263443 x 1807 x 43 x 7 x 3 x 2 + 1 =

10650056950807

[/size]

Sylvester's Sequence

[size=7]Several magical correspondence's
An infinite series of Egyptian fractions totaling 1
1=1/2+1/3+1/7+1/43+1/1807+1/3263443+1/xxxxx
Egyptian fractions are a fascinating nexus of history and number theory. It seems that Egyptian mathematics lacked the concept of irrational numbers (and 0) so to compensate they created a fractional system that consisted of sums of fractions with 1's in the numerator and different denominators

for instance 3/4 is not permitted, but 3/4=1/2+1/4 is
5/6 then 1/2+1/3
and so on
This sequence is known as the greedy Egyptian, that is it is an infinite series of Egyptian fractions equalling 1.

Sylvester's sequence then is exactly what Stormy quoted, which is a way to generate tis fraction sequence, which by algebraic manipulation is also
e(n)=[e(n-1)]^2 -e(n-1)+1
interestingly enough each of the denominators of the fractions is prime, and since it's an infinite series it demonstrates there is no greatest prime number[/size]
Rap

^^^^^^ see that up there.....

does that mean I'm reallllllllllllllllllly blonde

(blonde is good - it's nice, I can do that well!) (k - back in my box a?)

Lzzie:

k
well
1 + 2 = 3
3 x 2 + 1 = 7
7 x 3 x 2 + 1 = 43
43 x 7 x 3 x 2 + 1 = 1807
1807 x 43 x 7 x 3 x 2 + 1 = 3263443
3263443 x 1807 x 43 x 7 x 3 x 2 + 1 =

10650056950807


Rap:

Several magical correspondence's
An infinite series of Egyptian fractions totaling 1
1=1/2+1/3+1/7+1/43+1/1807+1/3263443+1/xxxxx
Egyptian fractions are a fascinating nexus of history and number theory. It seems that Egyptian mathematics lacked the concept of irrational numbers (and 0) so to compensate they created a fractional system that consisted of sums of fractions with 1's in the numerator and different denominators

for instance 3/4 is not permitted, but 3/4=1/2+1/4 is
5/6 then 1/2+1/3
and so on
This sequence is known as the greedy Egyptian, that is it is an infinite series of Egyptian fractions equalling 1.

Sylvester's sequence then is exactly what Stormy quoted, which is a way to generate tis fraction sequence, which by algebraic manipulation is also
e(n)=[e(n-1)]^2 -e(n-1)+1
interestingly enough each of the denominators of the fractions is prime, and since it's an infinite series it demonstrates there is no greatest prime number


Thank you guys; I can add nothing!