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Feel free to post questions in here, and there seems to be plenty of people looking to help!
Hi, I am offering anyone help on their Math studies. I am not the best mathematician to ever walk the face of the earth, nor am I the the coolest teacher you have ever know, but I am passionate about teaching, and would like to help anyone who needs it. I have a skype that you can call me at Trezeguet.math, and I will try to use http://www.dabbleboard.com/ to interact with people. I can help anyone up through some Calculus. I know that most people here are quite good at math and do not need help, so this may be a giant bomb, but I figure it can never hurt to offer. If you add me on skype, include something about TL in the invite bit and I'll add you and help you as soon as I can. I have no clue what the demand might be for this so I'll play it all by ear, so if you need some help, be it with 1 problem, or the whole kit and caboodle, PM me on TL or skype me! I think I can do Basic chemistry as well.
If anyone has ideas on how I can improve, swing away, I'm open to any idea.
Also, if you don't need help today, but anticipate needing even a little help later, PM what time generally works for you during the weekdays in US Central time. Eventually if there is enough demand, I'll try and set aside an hour or two every night to help people out.
 
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wow this is really nice.
I've seen several people post math help question blogs but now they can do it all here.
actually... schools about to start... how do you feel about thermodynamics?
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On September 03 2011 00:00 ComaDose wrote:
wow this is really nice.
I've seen several people post math help question blogs but now they can do it all here.
actually... schools about to start... how do you feel about thermodynamics?
I wish I knew more about thermo, but I only took an introductory course. I would guess that if you can handle whatever it was that got you to thermo, you will have at a minimum the knowledge that I have. Sorry! Edit: unless it's food thermodynamics, then I am more help...
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Chemistry conversions n stuff yo?
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On September 03 2011 00:05 SCDPetrify wrote:
Chemistry conversions n stuff yo?
I do some chemistry! You may just need to give me 5 minutes though to spruce up on whatever area you need help with.
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Awesome topic. And since I'm a complete moron when it comes to math, especially calculus, I may contact you in future.
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Theorem: 1$ = 1c
Proof:
1$ = 100c
1$ = (10c)2
1$ = (0.1$)2
1$ = 0.01$
1$ = 1c
wahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh~!
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On September 03 2011 00:45 rei wrote:
Theorem: 1$ = 1c
Proof:
1$ = 100c
1$ = (10c)2
1$ = (0.1$)2
1$ = 0.01$
1$ = 1c
wahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh~!
haha, well done. If you are so sure, I'll trade you unlimited times 1c for 1$.
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United States804 Posts
Holy shit, you're a hero... I'll add you later for some Calculus assistance :D
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How do you feel about binary?
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On September 03 2011 00:59 HackBenjamin wrote:
How do you feel about binary?
I don't really know what you mean, just doing basic math in base 2?
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On September 03 2011 00:50 Trezeguet wrote:Show nested quote +On September 03 2011 00:45 rei wrote:
Theorem: 1$ = 1c
Proof:
1$ = 100c
1$ = (10c)2
1$ = (0.1$)2
1$ = 0.01$
1$ = 1c
wahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh~! haha, well done. If you are so sure, I'll trade you unlimited times 1c for 1$.
Incorrect... you would have to have 100(c^2) to make that proof work.
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On September 03 2011 01:06 IntoTheheart wrote:Show nested quote +On September 03 2011 00:50 Trezeguet wrote:On September 03 2011 00:45 rei wrote:
Theorem: 1$ = 1c
Proof:
1$ = 100c
1$ = (10c)2
1$ = (0.1$)2
1$ = 0.01$
1$ = 1c
wahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh~! haha, well done. If you are so sure, I'll trade you unlimited times 1c for 1$. Incorrect... you would have to have 100(c^2) to make that proof work.
Thank you for the help, do you make square pennies by flattening them in molds on train tracks?
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On September 03 2011 01:18 Trezeguet wrote:Show nested quote +On September 03 2011 01:06 IntoTheheart wrote:On September 03 2011 00:50 Trezeguet wrote:On September 03 2011 00:45 rei wrote:
Theorem: 1$ = 1c
Proof:
1$ = 100c
1$ = (10c)2
1$ = (0.1$)2
1$ = 0.01$
1$ = 1c
wahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh~! haha, well done. If you are so sure, I'll trade you unlimited times 1c for 1$. Incorrect... you would have to have 100(c^2) to make that proof work. Thank you for the help, do you make square pennies by flattening them in molds on train tracks?
Well since you can't "square," the financial value of the penny without breaking the rule that 1$=100C, this sort of falls flat.
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Wow this is pretty cool man. I'll be sure to come here if my math is pwning my face :o
PS What did you vote for in this thread?
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I'm trying to find a string of positive integers that sum to 1000 and and who's product is as high as possible. I'm not sure how to approach it, my instinct is to say 2^500 is the answer but I'm pretty sure thats wrong. I simplified it to 10 and the answer is 2+2+3+3. I'm 99% sure it'll have to be a string of primes for sort of obvious reason that would be wordy for me to write out. I've been playing around with these two equations, x=p1^n1*p2^n2....pk^nJ where p is a prime and n is its power and p1(n1)+p2(n2)+...+pk(nj)=1000 where I limit the primes I use somewhat arbitrarily. If I could narrow it down to 2 and 3 or something like that I could optimize and be done... Oh well, hope some of that makes sense. Thanks for doing this
I don't want the answer but if you have a hint or idea I'd like to see it.
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On September 03 2011 01:28 Geovu wrote:Wow this is pretty cool man. I'll be sure to come here if my math is pwning my face :o PS What did you vote for in this thread?
I think I voted for 288 (mostly agreed to be correct) and 1/(2x). I regret that some people may see this as conflicting, but we aren't all perfect.
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On September 03 2011 01:36 n.DieJokes wrote:
I'm trying to find a string of positive integers that sum to 1000 and and who's product is as high as possible. I'm not sure how to approach it, my instinct is to say 2^500 is the answer but I'm pretty sure thats wrong. I simplified it to 10 and the answer is 2+2+3+3. I'm 99% sure it'll have to be a string of primes for sort of obvious reason that would be wordy for me to write out. I've been playing around with these two equations, x=p1^n1*p2^n2....pk^nJ where p is a prime and n is its power and p1(n1)+p2(n2)+...+pk(nj)=1000 where I limit the primes I use somewhat arbitrarily. If I could narrow it down to 2 and 3 or something like that I could optimize and be done... Oh well, hope some of that makes sense. Thanks for doing this
I don't want the answer but if you have a hint or idea I'd like to see it.
Ok, so I don't know the answer so I'm kind of typing out loud here, but to me 2^500 looks really good To me, it might be easier to solve it for 10, or 100 as the sum, and then look at how that will relate to 1000.
So for 10:
the possible integers are 1,2,3,4,5,6,7,8,9 and 10. Since we are trying to maximize the product, it is easy to say that 1 should be eliminated since that will not increase the product. Also, a simple answer is 4,6 and the product would be 24 so we know that a string of just "10" is wrong so we can eliminate that as well, also we know that 9 should not be used since the only combo is 1,9 which multiplies to 9. So this leaves us 2,3,4,5,6,7,8. Possibilities for 7 and 8 are
8,1,1 and 8,2 which (max = 16) doesn't beat 4,6
7,1,1,1 and 7,1,2 Max=(14)
so 7 and 8 are out
6,1,1,1,1 = 6
6,1,1,2 = 12
6,1,3 = 18
6,2,2 = 24
6,4 = 24
5,2,2,1 = 20
5,3,2 = 30
5,4,1 = 20
5,5 = 20
ok, so it's looking like more, smaller numbers tends to be better
4,2,2,2 = 32
4,4,2 = 32
4,3,3 = 36
even better
3,2,2,2,1 = 24
3,3,3,1 = 27
3,3,2,2 = 36
just as good
2,2,2,2,2 = 32
So looking at this, it makes me thing that perhaps the best answer isn't 2^500, but it is probably very close.
TLDR 2*2*(3^332) > 1*3^333 > 2^500
I just looked at 5^200 and (2^3)*(7^142) and it wasn't better. Also, obviously, using 4 and 2 are the same, but 2^3 > 8 and 2^4 > 16 so just use 2^x if you have extra bits left.
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Don't really have an answer and I'm not too good at math riddles, but I'm pretty sure your hunch about primes is correct:
If you have a non-prime number z in the answer, it can be written as z = x*y where at least one of x or y is prime. I don't have a proof for this next part but it seems apparent (by example) that x+y is always less than or equal to x*y. This means using a non-prime number is never "efficient" in terms of picking numbers for your set, and is often inefficient.
Might edit later with more stuff I notice while waiting for experiments to run at work :D
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Do you by any chance know quotient groups? I'm so confused about it... hahaha
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EPT
