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Mathematics Club

72,477 Views | 513 Replies
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Response to Mathematics Club 2004-04-04 09:50:28


At 4/1/04 05:20 PM, charleswhitesaber wrote: could someone help me with factoring trinomials

If you're still looking for help on this problem, please give me some more specific information on the kind of help you need.

Response to Mathematics Club 2004-04-06 07:04:42


Okay, here's a puzzle:
Farmer bob has 8 identical gold coins, and a uranium coin. The uranium coin is obviously waay heavier, but Farmer bob is suffering from radiation poisoning and really can't tell which one it is, by weight or any other indication. Farmer bob has a scale, and if two weights are placed on the scale, he can determine which side is heavier.

How many weighings does it take, and what process should he use to guarantee that he finds the uranium coin.

Response to Mathematics Club 2004-04-06 08:15:00


At 4/6/04 07:04 AM, Dr_Arbitrary wrote: Okay, here's a puzzle:

Assume Bob knows that u238 is heavier than Au

Table A is on right of scale
Table B is on left of scale
Tray C is off to the side somewhere

A = 2 coins + B = 2 coins + C = 3 coins

If A = lower, then u238 coin is in A
take coins from A, clear B and place 1 coin from A into A and one into B
Lower tray = u238 coin
(2 measurements)

If B = lower; repeat above process with coins from B
Lower tray = u238 coin
(2 measurements)

If A=B in levelness, C must contain u238, so
Clear A and B trays
Put one coin from C into A, out one coin into B and keep one in C
If A<>B then lower tray = u238
If A=B then C = u238
(3 measurements)

Response to Mathematics Club 2004-04-06 08:25:39


At 4/6/04 08:15 AM, River_Warden wrote: half-assed response

Sorry, last case is 4 remain in C
So must repeat final separate of contents of C up to 2 more times
(thus 3 measurements, but different description)

A1+A2 = B1+B2 (hold C1+C2+C3+C4)
C1+C2 <> C3+C4 (assume C1+C2 < C3+C4 for this demo)
C1 <> C2 (assume C1 < C2 for this demo)

Sorry for the premature release

Response to Mathematics Club 2004-04-06 08:49:47


I don't want to sound like a jerk, but from the looks of it, your answer is incorrect. I think you were confused by the wording of my question, sorry about that.

Just to clarify, there are 9 coins total, 8 of them are actually gold.

Response to Mathematics Club 2004-04-06 13:23:00


Im brilliant. I taught myself how to transform something from a region to another, with Jacobian, and I found a better way than my Professor. Im pimp dizzle.

Response to Mathematics Club 2004-04-06 16:40:36


Hey everybody, I'm pretty sure I've got mathematica working for good, so if anyone has anything extremely difficult they need worked out, just let me know. Mathematica -> Power

Response to Mathematics Club 2004-04-06 23:44:30


At 4/6/04 08:49 AM, Dr_Arbitrary wrote: something far too polite

You sir, are too kind. The fur of the ass shall be worn proudly upon my shoulders both for reading comprehension and quick trigger finger. (head hanging a little lower).

If I may try again, (same assumption that the farmer recognizes that the u238 coin will weigh more than the gold coin)

Tray on Left of Balance = A
Tray on Right of Balance = B
Tray on table beside Balance = C

First Weighing
A1+A2+A3 vs B1+B2+B3 (C1+C2+C3 set aside)

Condition 1
A = B; therefore u238 is contained within set C
Clear A and B and distribute 1 coin from C into each tray

Condition 1.1
A=B thus C is the u238 coin
Condition 1.2
A>B thus A is the u238 coin
Condition 1.3
A<B thus B is the u238 coin

Total # of Measurements for Condition 1 = 2

Condition 2
A>B
Clear tray B and C (pocket the gold) and distribute 1 coin from set A into each tray (A, B and C)

Condition 2.1
A=B thus C is the u239 coin
Condition 2.2
A>B thus A is the u238 coin
Condition 2.3
A<B thus B is the u238 coin

Total # of Measurements for Condition 2 = 2

Condition 3
A<B
Clear tray A and C (pocket the gold) and distribute 1 coin from set B into each tray (A, B and C)

Condition 3.1
A=B thus C is the u239 coin
Condition 2.2
A>B thus A is the u238 coin
Condition 2.3
A<B thus B is the u238 coin

Total # of Measurements for Condition 3 = 2

All Condition Scenarios require no less than 2 comparisons to determine the u238 coin

Response to Mathematics Club 2004-04-07 00:49:16


The winner, it's you, River_Warden.

I'm a am drunk right now, so I'll leave you with another puzzle if you're up for it,
part 1, what is a general equation for the sum of consecutive integers from 1 to x.
basically, what equation will fulfill the equation, 1,3,6,10,15...

Part 2, Instead of looking up an answer, figure it out yourself and describe the techinique you used.

Part 3, try it with consecutive square numbers, or cubes

Response to Mathematics Club 2004-04-07 01:57:06


At 4/7/04 12:49 AM, Dr_Arbitrary wrote: part 1, what is a general equation for the sum of consecutive integers from 1 to x.
basically, what equation will fulfill the equation, 1,3,6,10,15...

Part 2, Instead of looking up an answer, figure it out yourself and describe the techinique you used.

Part 3, try it with consecutive square numbers, or cubes

Part 1 - have long since forgotten the equation 8-(
Ok, lets see....

Don't know how to so sigma so I'll just write SIG

X is the maximum variable on top of SIG
I is equal to 1 (as a start, because my first thought is -Integers and +Integers are going to behave diffently unless you use ABS())

Start with 1 and add X, times the maximum number X then divide the pair (2) (the logic is breaking down here but I'll try to explain what I am thinking - may not be right)

Start with 1
If I count 1 of something (like fingers) and assign the ordinal value to each finger, then mirror to the other hand and inverse it

1 on bottom, 1 on top, the sum the pair is 2 = (x + 1)
2 x 1 =2
Divide 2 by 2 = 1

Next x is 2
If I count 2 fingers, then mirror to the other hand and inverse it

1 2 on bottom, 2 1 on top, the sum of any pair is 3 (x + 1)
3 x 2 =6
Divide 6 by 2 = 3

Say x is 3
If I count 3 fingers, then mirror to the other hand and inverse it

1 2 3 on bottom, 3 2 1 on top, the sum of any pair is 4 (x + 1)
4 x 3 =12
Divide 12 by 2 = 6

And so on so the progression is 1, 3, 6, 10, 15 etc...

I know, solve for x with fingers.....pretty lame

Anyway, back to the thought process...

If I write the Integers down in a column and match the running sum beside it like so, I see the following relationship:

Int Sum Delta

1 1
2 3 X^2 - 1(X-1)
3 6 X^2 - 1X
4 10 X^2 - 2(X-1)
5 15 X^2 - 2X
6 21 X^2 - 3(X-1)
7 28 X^2 - 3X
8 36 X^2 - 4(X-1)
9 45 X^2 - 4X
10 55 X^2 - 5(X-1)

Part 2/3 - Oooow my head hurts.

At this point, I have to take a break and actually do some work related work.

Please critique the method (madness) so far and let me know where I need to catch the train to get back on track

Response to Mathematics Club 2004-04-07 03:45:34


Somehow, I'm still drunk, but I'll give you a hint, the generalized formula for the sum of x for x=1 to n =
1/2 * n * (n+1)... I think

Response to Mathematics Club 2004-04-07 05:01:41


At 4/7/04 03:45 AM, Dr_Arbitrary wrote: Somehow, I'm still drunk, but I'll give you a hint, the generalized formula for the sum of x for x=1 to n =
1/2 * n * (n+1)... I think

Answer to part 1 - yes

Now the extrapolation back onto the finger scenario for the single instance (calculate the next number)

n = set of evens
n+1 = set of odds
1/2 is the normalization of the delta

I'll try to draw it out:

87654321
++++++++
12345678
---------------
99999999

9 * 8 = 72
(n+1) * n = 72

72 / 2 = 36
1/2 * n * (n+1)

2x Check against series => 1,3,6,10,15,21,28,...36

So the first series is on track, on to the Squares and Cubes

Response to Mathematics Club 2004-04-07 11:15:29


I'm not going to comment on your math so far, mostly because I'm really tired and am in no position to critique anything.

There's really no 'correct' answer for the problem I posed, I just remember how I did it, and wondered if anyone else had a creative way of figuring it out. So far, your answer looks VERY creative.

Response to Mathematics Club 2004-04-07 13:10:28


At 4/7/04 11:15 AM, Dr_Arbitrary wrote: I'm not going to comment on your math so far, mostly because I'm really tired and am in no position to critique anything.

There's really no 'correct' answer for the problem I posed, I just remember how I did it, and wondered if anyone else had a creative way of figuring it out. So far, your answer looks VERY creative.

You definitely have a future in politics, if you so desire. Numbers and I have never officially been friends. I score better with symbols and occasionally recognize patterns in development.
As for the rote memorization 8-P. Too tired at this point to review further but will approach fresh in the next 18-20 hours after some sleep. The second post is way out. Need to rethink it. Currently looking at a series of structured arrays of macaroni 1x2, 2x3, 3x4, 4x5 (helps with the visualization)

- side point - can't remember where I picked up the finger method but seem to recall an application where the consecutive integers didn't need to start at 1 .

I'll let ya know if I come up with something useful. General question, outside of the direct solution to the formula, is there a practical application for running series

i.e. Parabolic curves (quadratic functions) describe interaction of moving bodies under the influence of gravity

Linear equations describe an ideal vector without influence of outside forces

Derivatives describe the area under a curve - hence volume of a non-linear geometric shape

Closest thing I could come up with would be propagation of something (objects = physical volume, or something abstract such as $ = a bank balance) with uniform growth

Sorry if the posts are a little abstract. I am not currently engaged with a specific course and am approaching this as a refresher to firm up the grey matter

Response to Mathematics Club 2004-04-08 22:09:08


Hey I'm pretty good at math. I'm kinda accelerated about 3 years beyond my peers, so that's not bad. I can basically answer any question up to some parts of Calculus.

AIM id: butterballlz

Response to Mathematics Club 2004-04-11 19:19:24


At 4/8/04 10:09 PM, Ravens_Grin wrote: Hey I'm pretty good at math. I'm kinda accelerated about 3 years beyond my peers, so that's not bad. I can basically answer any question up to some parts of Calculus.

Sure, go ahead and post. If you have any questions feel free to ask because there are a few of us that can do high level math.

Which reminds me, Dr. A, do you know how to use Mathematica for Vector Fields? If so, I may ask for a few VF screen shots.

Response to Mathematics Club 2004-04-11 19:43:34


I'm having uber trouble with my math homework... 2... +...2....=??? Oh dear God it's too hard I'm going to kill myself!! The frustration!!

Response to Mathematics Club 2004-04-11 20:07:33


At 4/11/04 07:43 PM, TheGrayFox wrote: I'm having uber trouble with my math homework... 2... +...2....=??? Oh dear God it's too hard I'm going to kill myself!! The frustration!!

I know that youre being sarcastic, and therefore you intended to be funny, but that wasnt funny at all. Try again.

Response to Mathematics Club 2004-04-11 20:42:06


Fuck I would never be able to do any of this shit. I will just stick to my social sciences. Oh God I am dreading takign economics already.


Bellum omnium contra omnes

BBS Signature

Response to Mathematics Club 2004-04-11 20:47:22


At 4/11/04 08:42 PM, RugbyMacDaddy wrote: Fuck I would never be able to do any of this shit. I will just stick to my social sciences. Oh God I am dreading takign economics already.

Economics only uses very basic algebra for Macro and Micro 1. Once you get to higher level Econ classes youll use Calculus, but the Calculus youll use is very basic as well. Unless youre taking Econometrics, you dont need to worry about the mathematical portion of the class.

Response to Mathematics Club 2004-04-11 21:47:34


At 4/11/04 07:19 PM, BWS wrote: Which reminds me, Dr. A, do you know how to use Mathematica for Vector Fields? If so, I may ask for a few VF screen shots.

I know what you're asking, although it's been a while since I had to do any vector fields. I have no clue how to get mathematica to do what you want. I really only use it as a calculator.

Response to Mathematics Club 2004-04-11 22:04:47


At 4/11/04 08:07 PM, BWS wrote:
I know that youre being sarcastic, and therefore you intended to be funny, but that wasnt funny at all. Try again.

You didn't find that funny because you're a gay ass math nerd. You're a "Cmon let's do calculus" on a Saturday party night kind of guy, sad.

Response to Mathematics Club 2004-04-11 22:46:43


Here's a "new math" problem:
Jimmy has 6 crayons. 4 of them are broken.
Suzie has 8 crayons. 6 of them are broken.
Who should be sadder?


Jack of all trades, Master of none

Head Scientist of NG Mafia

& proud member of Anime Club

BBS Signature

Response to Mathematics Club 2004-04-11 22:58:00


At 4/11/04 10:46 PM, mr_trivia wrote: Here's a "new math" problem:
Jimmy has 6 crayons. 4 of them are broken.
Suzie has 8 crayons. 6 of them are broken.
Who should be sadder?

The crayons because they're broken :( Poor crayons....

I think the correct answer is... Neither because they share the same value of crayons... Or... Synagogue..

Response to Mathematics Club 2004-04-11 23:10:22


At 4/11/04 10:58 PM, TheGrayFox wrote:
At 4/11/04 10:46 PM, mr_trivia wrote: Here's a "new math" problem:
Jimmy has 6 crayons. 4 of them are broken.
Suzie has 8 crayons. 6 of them are broken.
Who should be sadder?
The crayons because they're broken :( Poor crayons....

I think the correct answer is... Neither because they share the same value of crayons... Or... Synagogue..

That's the thing about new math: there's no wrong answer. It's supposed to build children's self-esteem and disguise the fact that they're dumbasses.


Jack of all trades, Master of none

Head Scientist of NG Mafia

& proud member of Anime Club

BBS Signature

Response to Mathematics Club 2004-04-11 23:40:57


At 4/11/04 11:10 PM, mr_trivia wrote:
That's the thing about new math: there's no wrong answer. It's supposed to build children's self-esteem and disguise the fact that they're dumbasses.

Hahaha, exactly... When my father was studying at a European University, if you were caught using a calculator you would honestly get beat and kicked out of class for a week. Crazy :) He's amazing at math, chemistry and physics sigh would I kill to be as smart as him. He laughs at all the “New Math” he thinks it’s for morons “Color in ½ of a circle” Congratulations… You’re a moron. Nowadays… The best way to learn is in your room, by yourself.

Response to Mathematics Club 2004-04-12 00:44:35


My personal recommendation is Schaums Outlines, they have tons of worked problems on most subjects. Right now I'm learning group theory.

I wonder how much money it would cost for the Department of Education to pay for an internet course that anyone could take, not for credit or anything, maybe a standardized test for that, but courses so that anyone who wants to learn algebra can learn it from the internet.

I know that the US Navy has a program that pretty much is independent of instructors to teach the fundamentals of electronics. They use a program called NIDA, it worked pretty well for me and most people really only need help with a few tricky parts.

Response to Mathematics Club 2004-04-12 05:28:56


Another good way to teach Math.... Fear...

Response to Mathematics Club 2004-04-12 06:43:17


If I ever become a high school teacher, I'll have a book with names of all the students, and a stopwatch. Every once in a while I'll point out a student and ask them a question, then turn on the stopwatch. Then, no matter how quickly they get the answer, I'll shake my head disaprovingly and jot something down in the book.

Response to Mathematics Club 2004-04-12 07:02:01


At 4/2/04 01:19 AM, RotesStinktier wrote: I was just wondering if anyone here had heard of / is participating in The Great Internet Mersenne Prime Search?

Late reply, but I've ran that for about three years. Not running it at the moment though as I've not been bothered to install it on the new computer. Let's check my stats:

LL tests: 1.728 years, 7 done
Factoring: 2.179 years, 7 found

And quickly using my other computer has assigned itself a new factoring job. Shame that computer's never on and the assignment will expire.

Ooh, that factoring's still keeping me in the top 1000, a mild surprise.