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Author Topic:   give me an algerbra problem
geekwannabebritt
Geek Apprentice

Posts: 41
From: Washington
Registered: Jul 2002

posted August 12, 2002 23:15     Click Here to See the Profile for geekwannabebritt   Click Here to Email geekwannabebritt     Edit/Delete Message   Reply w/Quote
i want to see if i can solve it

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Drazgal
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Posts: 35
From: Cleveland, United Kingdom
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posted August 13, 2002 04:03     Click Here to See the Profile for Drazgal   Click Here to Email Drazgal     Edit/Delete Message   Reply w/Quote
Heres a fun one from my A Level revision notes

Solve the differential eq^n

y * cotx * dy/dx = 2 at x = Pi/4 and y = 0

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Alien Investor
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From: New York City
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posted August 13, 2002 07:03     Click Here to See the Profile for Alien Investor   Click Here to Email Alien Investor     Edit/Delete Message   Reply w/Quote
Drazgal, your problem is 2nd year calculus, not algebra.

(1) A Manhattan resident commutes back and forth to work for a total of two trips per day. An ordinary commuting week is five days. It costs $1.50 to ride the subway or $6.00 to take a taxicab. The Manhattanite spent $28.50 on commuting this week. How many subway trips and how many taxicab trips did they take?

To solve this properly, use variables (x = # of subway trips, y = # of taxi trips) and set up two equations in two unknowns. Then reduce to 1 equation in one unknown.

(2) Find the sum of the integers from 1 to 100, inclusive (this is not quite algebra). Then find a more general formula for the sum of the integers from 1 to N (this is algebra).

(3) Find a solution to the general quadratic equation, a x^2 + b x + c = 0. It's okay to use books and the Internet to search. Study the solution until you can do it yourself from memory. Then try it with different letters and different signs all the way through, such as k y^2 - m y + n = 0. This will give you plenty of practice in pushing equations around.

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nekomatic
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Posts: 452
From: Manchester, UK
Registered: Mar 2000

posted August 13, 2002 07:25     Click Here to See the Profile for nekomatic   Click Here to Email nekomatic     Edit/Delete Message   Reply w/Quote
(4) Let x be 2. If it tries to be anything else, prevent it. Use force if necessary.

(5) If m is the co-dependant of n, n is fixated upon half of y and z is a member of a triangle involving m and q, what is the solution and should y be held responsible?

(6) Find the cube root of b. I can't remember where I put it and I really need it for tomorrow.

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spungo
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From: Hell's toilet
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posted August 13, 2002 07:30     Click Here to See the Profile for spungo     Edit/Delete Message   Reply w/Quote
quote:
Originally posted by Alien Investor:
Drazgal, your problem is 2nd year calculus, not algebra.

Err... that's high school calculus. Oh - and the answer is

y = i*SQRT(2*ln(2*(cos^2(x))))

(sorry... the geek in me can never resist!)

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"No longer empty and frantic like a cat tied to a stick, that's driven into frozen winter shit (the ability to laugh at weakness), calm, fitter, healthier and more productive, a pig in a cage on antibiotics."

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maxomai
Super Geek

Posts: 248
From: Portland, OR
Registered: May 2001

posted August 13, 2002 07:51     Click Here to See the Profile for maxomai   Click Here to Email maxomai     Edit/Delete Message   Reply w/Quote
Easy: Show that Rubik's Cube is abelian. (This is an edit; the original was wrong)

Medium: Prove that it is impossible to square the circle (that is, given a circle of a given area, to produce a square of exactly that area) using a straightedge and a compass.

Hard: Find a general algorithm for generating the Galois groups for C^n as an extension field of R^n.

(Yes, these are algebra problems, and they are solvable. And yes, I'm a sadistic jerk.)

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+Andrew
Alpha Geek

Posts: 275
From: Boston, MA, USA
Registered: Aug 2001

posted August 13, 2002 09:02     Click Here to See the Profile for +Andrew   Click Here to Email +Andrew     Edit/Delete Message   Reply w/Quote
(1) 7 subway, 3 taxi

(2) 5050. Formula: (1+N)*(N/2) (A famous problem)

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Alien Investor
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Posts: 488
From: New York City
Registered: Jan 2000

posted August 13, 2002 09:58     Click Here to See the Profile for Alien Investor   Click Here to Email Alien Investor     Edit/Delete Message   Reply w/Quote
quote:
Originally posted by spungo:
y = i*SQRT(2*ln(2*(cos^2(x))))


This looks like it came out of a program, what with the redundant instances of 2 in "2 * ln(2 * ...)", and especially multiplying i by the square root of the logarithm of a number in the range [0..1].

That is to say: cos^2(x) is in [0,1]; so ln cos^2(x) is negative; so sqrt(2*ln(2*cos^2(x))) is pure imaginary, and i times that mess is real again.

Try solving by hand and the redundant 2's come out and a "-4 ln cos x" is really "4 ln sec x" and stays purely real.

Deep breath ...

y * cot x * dy/dx = 2

y dy = 2 tan x dx

1/2 y^2 = 2 ln sec x + C1

y^2 = 4 ln sec X + C2 # Constant-of-integration fiddling

y = +2 sqrt ln sec X + C3
y = -2 sqrt ln sec X + C4
# twin solutions

Plug the initial conditions and solve for C3, C4:

0 = +2 sqrt ln sec pi/4 + C3
0 = +2 sqrt ln sqrt 2 + C3
0 = +2 sqrt ((ln 2)/2) + C3
0 = sqrt(ln 8) + C3
C3 = - sqrt ln 4

Similarly,

0 = -2 sqrt ln sec pi/4 + C4
0 = -2 sqrt ln sqrt 2 + C4
0 = -2 sqrt ((ln 2/2) + C4
0 = -sqrt(ln 4) + C4
c4 = sqrt ln 4
[/i]

(Note I am using sqrt here to mean "positive square root" always, consistently.)

So:


y = +2 sqrt ln sec x - sqrt ln 4
y = -2 sqrt ln sec x + sqrt ln 4

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Alien Investor
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From: New York City
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posted August 13, 2002 10:08     Click Here to See the Profile for Alien Investor   Click Here to Email Alien Investor     Edit/Delete Message   Reply w/Quote
quote:
Originally posted by maxomai:
Easy: Show that Rubik's Cube is abelian. ([b]This is an edit; the original was wrong)
[/B]

I think the Cube is non-Abelian. For example, turn the top face one quarter turn to the right, then turn the right face one quarter turn back. Those two operations do not commute. (Follow the top left front corner cubie).

I understand the Medium problem, but I can't solve the whole problem. First, show that the compass and straight-edge can perform a subset of the rational functions (addition, subtraction, multiplication, division, square roots, nth roots). That's not hard. Then, prove that pi is transcendental. That's beyond me, but I remember where to find the proof in Spivak.

I don't even understand the Hard problem, my group theory is gone with the wind.

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maxomai
Super Geek

Posts: 248
From: Portland, OR
Registered: May 2001

posted August 13, 2002 10:27     Click Here to See the Profile for maxomai   Click Here to Email maxomai     Edit/Delete Message   Reply w/Quote
quote:
Originally posted by Alien Investor:
I think the Cube is non-Abelian. For example, turn the top face one quarter turn to the right, then turn the right face one quarter turn back. Those two operations do not commute. (Follow the top left front corner cubie).

I should have listened to my first instinct. You're right, of course.

quote:

I understand the Medium problem, but I can't solve the whole problem. First, show that the compass and straight-edge can perform a subset of the rational functions (addition, subtraction, multiplication, division, square roots, nth roots). That's not hard. Then, prove that pi is transcendental. That's beyond me, but I remember where to find the proof in Spivak.

That's about the gist of it although if I recall correctly cube roots are impossible .. that would be the doubling-the-cube problem.

quote:

I don't even understand the Hard problem, my group theory is gone with the wind.

It's pretty easy as it turns out, but it took me forever to figure it out when I was a grad student. Chalk it up to my being an idiot. Anyways:

Suppose n = 1. The Galois group Gal(C,R) consists of identity (a+bi => a+bi) and conjugate (a+bi => a-bi). This makes sense since [C:R] = 2. Note that there are no intermediate splitting fields since [C:R] is prime.

Suppose n = N > 1. C^n is an extension field of C^(n-1) * R (up to isomorphism). Furthermore Gal (C^n, C^(n-1) * R) = {phi1, phi2} such that:

phi1 is identity;
phi2 is identity on all but one dimension of C^n, and is conjugate for the one exception.

Note that Gal(C^(n-1) * R, C^(n-2) * R^2) is isomorphic to Gal (C^(n-1), C^(n-2) * R). Thus the induction hypothesis applies.

I admit to some hand waving and detail skipping here.

This should help you follow along: http://www.math.niu.edu/~beachy/aaol/galois.html

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Xanthine
Uber Geek

Posts: 815
From: the lab
Registered: Mar 2001

posted August 13, 2002 10:29     Click Here to See the Profile for Xanthine     Edit/Delete Message   Reply w/Quote
quote:
Originally posted by maxomai:
Easy: Show that Rubik's Cube is abelian. ([b]This is an edit; the original was wrong)

Medium: Prove that it is impossible to square the circle (that is, given a circle of a given area, to produce a square of exactly that area) using a straightedge and a compass.

Hard: Find a general algorithm for generating the Galois groups for C^n as an extension field of R^n.

(Yes, these are algebra problems, and they are solvable. And yes, I'm a sadistic jerk.)

[/B]


I agree. After a year and a half of college calculus I don't even understand what the easy one is asking.

Here's a relatively easy one for her.
This is the concentration:volume relationship used for diluting solutions in chemistry: initial concentration(initial volume) = final concentration(final volume). You've got a stock solution of 12 M hydrochloric acid (pour carefully). You want to make a 50 mL, 2 M solution (M represents the units of concentration). How much 12 M stock and how much water are you going to need?

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Take by surprise and the world gives up resistance.
- Tennesee Williams

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maxomai
Super Geek

Posts: 248
From: Portland, OR
Registered: May 2001

posted August 13, 2002 11:06     Click Here to See the Profile for maxomai   Click Here to Email maxomai     Edit/Delete Message   Reply w/Quote
quote:
Originally posted by Xanthine:
I agree. After a year and a half of college calculus I don't even understand what the easy one is asking.

If you're a math major you might understand all three questions by your Senior year.

(Yes, I'm sick .. sick sick sick.)

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SpikeSpiegel
Highlie

Posts: 739
From: my chair
Registered: Jun 2002

posted August 13, 2002 11:08     Click Here to See the Profile for SpikeSpiegel   Click Here to Email SpikeSpiegel     Edit/Delete Message   Reply w/Quote
quote:
Originally posted by Alien Investor:
Drazgal, your problem is 2nd year calculus, not algebra.



oh so thats why i couldnt do it, here i thought i was just losing my mind

------------------
"The Borg? sounds swedish"

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neotatsu
Uber Geek

Posts: 839
From: A place my soul no longer resides
Registered: Jun 2002

posted August 13, 2002 15:53     Click Here to See the Profile for neotatsu   Click Here to Email neotatsu     Edit/Delete Message   Reply w/Quote
And to think, this thread was started by a high school FRESHMAN to see if she could do SIMPLE ALGEBRA, and because I'm in a pissy mood, I'm not going to post any either(though I'm seriously tempted to write a few and send them out in an e-mail, though I don't think that'd be a good idea, Twinkie only has access to a public terminal I believe..)

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skylar
Alpha Geek

Posts: 342
From: the creepy house on the corner
Registered: May 2002

posted August 13, 2002 17:50     Click Here to See the Profile for skylar   Click Here to Email skylar     Edit/Delete Message   Reply w/Quote
I was just thinking it's a shame Twink's not here to read this. This is definitely her kind of thread...

sky

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Alien Investor
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Posts: 488
From: New York City
Registered: Jan 2000

posted August 13, 2002 23:22     Click Here to See the Profile for Alien Investor   Click Here to Email Alien Investor     Edit/Delete Message   Reply w/Quote
Just to let Britt in on the joke ...

There are two different branches of math, both named algebra. The first "algebra" starts with simple word problems like "how often does the commuter take the subway" and goes up to, more or less, solving polynomial equations. Americans usually learn this algebra in 7th or 8th or 9th or 10th grade. I can just feel the geek one-upmanship reflex kicking in among all the posters here, so I'll say that I started learning this style of "algebra" in the summer between 2nd and 3rd grade.

The second "algebra" starts with generalizations over all the possible varieties of the first "algebra". It starts with matrices (linear algebra) and includes group theory (abstract algebra) and plenty of other stuff that I've either forgotten or never knew. American geeks usually learn this algebra starting around the second year of college.

There's a large gap between "algebra" (type 1) and "algebra" (type 2).

I'm assuming from Britt's school level (starting high school) that she is interested in the first type of algebra, and the joke is that people are posting type-2 algebra problems. Which is fine with me, I love math.

And it would be even funnier if Britt were the second incarnation of Ramanujan and she really is looking for problems like the Goldbach conjecture (every even number > 2 is the sum of two primes), which is my favorite unsolved problem.

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greycat
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posted August 14, 2002 13:13     Click Here to See the Profile for greycat     Edit/Delete Message   Reply w/Quote
Here's a nice, easy one that a high school student should be able to handle:

Pick any number, N. Square it and subtract 1. Remember the answer. Now take N+1 and N-1 and multiply them. Does the answer look familiar?
Why? Either prove that the two answers are always the same for all N, or find a counterexample N for which they are different.

Here's a grunt-work one (not a lot of thought involved, just technique):

Find the solutions to the equation x^3 + x^2 - 4x - 24 = 0. (Hint: one is real, and two are complex. Once you find the real root, you can get the other two with the quadratic formula [Alien Investor problem #3].)

Here's a much harder one:

Prove that there are infinitely many primes. (There are many famous proofs of this; if you get one from a book or the Internet, you need to make sure you understand it completely and can reproduce it from memory.)

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spungo
Highlie

Posts: 749
From: Hell's toilet
Registered: Jan 2002

posted August 15, 2002 02:01     Click Here to See the Profile for spungo     Edit/Delete Message   Reply w/Quote
quote:
Originally posted by Alien Investor:
This looks like it came out of a program, what with the redundant instances of 2 in "2 * ln(2 * ...)", and especially multiplying i by the square root of the logarithm of a number in the range [0..1].
[/i]

Err... no, it was done by hand - you may dislike the style, but I thought it was a more compact expression. I always thought if one can be express it conveniently in one term (instead of two) than one should do so.

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"No longer empty and frantic like a cat tied to a stick, that's driven into frozen winter shit (the ability to laugh at weakness), calm, fitter, healthier and more productive, a pig in a cage on antibiotics."

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snupy
Highlie

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From: Chicago
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posted August 15, 2002 03:10     Click Here to See the Profile for snupy   Click Here to Email snupy     Edit/Delete Message   Reply w/Quote
quote:
Originally posted by spungo:
Err... that's high school calculus. Oh - and the answer is

y = i*SQRT(2*ln(2*(cos^2(x))))

(sorry... the geek in me can never resist!)


**whew** lucky me-I learned my lesson last time and brought a change of underpants.

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Alien Investor
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From: New York City
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posted August 15, 2002 06:53     Click Here to See the Profile for Alien Investor   Click Here to Email Alien Investor     Edit/Delete Message   Reply w/Quote
quote:
Originally posted by spungo:
Err... no, it was done by hand - you may dislike the style, but I thought it was a more compact expression. I always thought if one can be express it conveniently in one term (instead of two) than one should do so.


There's an advantage to 1 term instead of 2, but you still missed the point about generating imaginary intermediate results. If nothing else, on your terms, that's some redundant multiplication that can be folded together.

Viz:

y = i sqrt (2 * ln (2 * cos^2(x)))

Versus:

y = sqrt (-2 * ln (2 * cos^2(x)))

And then:

y = sqrt (2 * ln (1/2 * sec^2(x)))

y = sqrt (ln (1/4 * sec^4(x)))

Cheers.


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spungo
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posted August 15, 2002 07:16     Click Here to See the Profile for spungo     Edit/Delete Message   Reply w/Quote
Ok, fair enough - but I disagree that leaving the 'i' on the outside is implicitly wrong - it's not as if the argument to the sqrt will always be negative.

quote:
From Alien Investor:
That is to say: cos^2(x) is in [0,1]; so ln cos^2(x) is negative; so sqrt(2*ln(2*cos^2(x))) is pure imaginary,

Not true!
sqrt(2*ln(2*cos^2(x))) is NOT pure imaginary! 2*cos^2(x) is NOT always between 0 and 1.

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"No longer empty and frantic like a cat tied to a stick, that's driven into frozen winter shit (the ability to laugh at weakness), calm, fitter, healthier and more productive, a pig in a cage on antibiotics."

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maxomai
Super Geek

Posts: 248
From: Portland, OR
Registered: May 2001

posted August 15, 2002 08:38     Click Here to See the Profile for maxomai   Click Here to Email maxomai     Edit/Delete Message   Reply w/Quote
quote:
Originally posted by greycat:
Here's a much harder one:

Prove that there are infinitely many primes. (There are many famous proofs of this; if you get one from a book or the Internet, you need to make sure you understand it completely and can reproduce it from memory.)


If you like that one:

A Smith number is a positive composite integer, such that the sum of its digits, is equal to the sum of the digits of its prime factors.

Examples:

4 is a Smith number (prime factors are 2 and 2. 2+2 = 4)
22 is a Smith number (prime factors are 2 and 11. 2+1+1 = 4)

23 is not a Smith number because it is prime.
26 is not a Smith number. (prime factors are 13 and 2. 1+3+2 = 6; 2+6 = 8)

666 is a Smith number. (prime factors are 2, 3, 3, 37. 2+3+3+3+7 = 18; 6+6+6 = 18)

Prove that there are infinitely many Smith numbers. (This has been done before)

Bonus question: prove that for each K, there are infinitely many positive composite integers so that K divides both the sum of the digits and the sum of the digits of the prime factors. (This has been done before.)

Super duper bonus question: prove that there is a prime number between every two squares (e.g., 2 is between 1 and 4; 5 is between 4 and 9; 11 is between 9 and 16; etc.) This is called Sierpinski's Conjecture and it hasn't been proven yet, despite the efforts of a bunch of math geeks one weekend at Purdue University. The smartest of the bunch thinks the solution is algebraic (that's "type 2" algebra).

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Alien Investor
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posted August 15, 2002 08:41     Click Here to See the Profile for Alien Investor   Click Here to Email Alien Investor     Edit/Delete Message   Reply w/Quote
The cool thing about Geek Culture is that nobody will dare yell at us for being too geeky in a public place.

You are right! 2 cos^2(x) is in the range [0,2], not [0,1]. I dropped the "2 *" part in my head when I was analyzing it. Oops.

Just looking at the equation again:

y * cot(x) * dy/dx = 2

The "y * dy/dx" term implies that there will be two mirror-image solutions, with y = f(x) and y = -f(x).

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spungo
Highlie

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From: Hell's toilet
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posted August 15, 2002 08:47     Click Here to See the Profile for spungo     Edit/Delete Message   Reply w/Quote
quote:
Originally posted by Alien Investor:
The cool thing about Geek Culture is that nobody will dare yell at us for being too geeky in a public place.


I know! It's a place we can all be ourselves - we need to arrange some Geek Pride marches, or something!

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"No longer empty and frantic like a cat tied to a stick, that's driven into frozen winter shit (the ability to laugh at weakness), calm, fitter, healthier and more productive, a pig in a cage on antibiotics."

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snupy
Highlie

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From: Chicago
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posted August 15, 2002 09:40     Click Here to See the Profile for snupy   Click Here to Email snupy     Edit/Delete Message   Reply w/Quote
It's kinda like online porn for me, given what geekiness does to me...

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spungo
Highlie

Posts: 749
From: Hell's toilet
Registered: Jan 2002

posted August 15, 2002 09:51     Click Here to See the Profile for spungo     Edit/Delete Message   Reply w/Quote
quote:
Originally posted by snupy:
It's kinda like online porn for me, given what geekiness does to me...

Online porn for me would be... I suppose... a beer forum... or a beer-and-doughnut forum.... (mmmmm.... doughnuts!...) **tremble**

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"No longer empty and frantic like a cat tied to a stick, that's driven into frozen winter shit (the ability to laugh at weakness), calm, fitter, healthier and more productive, a pig in a cage on antibiotics."

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Alien Investor
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From: New York City
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posted August 15, 2002 11:18     Click Here to See the Profile for Alien Investor   Click Here to Email Alien Investor     Edit/Delete Message   Reply w/Quote
quote:
Originally posted by snupy:
It's kinda like online porn for me, given what geekiness does to me...

New York Apartment Porn:

"Now I'm walking into your ENORMOUS living room ..."

I swear, there are days I would pay $2 a minute to hear something like that.

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snupy
Highlie

Posts: 763
From: Chicago
Registered: Mar 2002

posted August 15, 2002 11:41     Click Here to See the Profile for snupy   Click Here to Email snupy     Edit/Delete Message   Reply w/Quote
quote:
Originally posted by Alien Investor:
New York Apartment Porn:

"Now I'm walking into your ENORMOUS living room ..."

I swear, there are days I would pay $2 a minute to hear something like that.


ROTFLOL!!!!!!!!!

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maxomai
Super Geek

Posts: 248
From: Portland, OR
Registered: May 2001

posted August 15, 2002 12:40     Click Here to See the Profile for maxomai   Click Here to Email maxomai     Edit/Delete Message   Reply w/Quote
quote:
Originally posted by Alien Investor:
New York Apartment Porn:

"Now I'm walking into your ENORMOUS living room ..."

I swear, there are days I would pay $2 a minute to hear something like that.


Which, by coincidence, is almost as much as it costs to rent in Lower Manhattan

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