Maths- Gym for Brainiacs

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Post by Babun Mon Apr 01, 2013 11:20 am

kiranr wrote:
I think this is about arranging each column in a sequence.

So for 2 columns, there are 2! = 2 ways. For 3 columns, there are 3! = 6 ways. For 4 columns, there are 4! = 24 ways. For 6 columns there are 6! = 720 ways. For n columns, there are n! ways.
spanky wrote:
kiranr wrote:
I think this is about arranging each column in a sequence.

So for 2 columns, there are 2! = 2 ways. For 3 columns, there are 3! = 6 ways. For 4 columns, there are 4! = 24 ways. For 6 columns there are 6! = 720 ways. For n columns, there are n! ways.

im gonna sign this post with a gif

Maths- Gym for Brainiacs  - Page 27 B7H1K
You didn't think the question would be this easy? Am I known for easy questions? eco smile
alexander mahone wrote:Is it really that many ways for 4 columns? I don't see it, lol, maybe I'm seeing it wrong. I tried to just expanding from the result for 3 columns to get the numbers of ways for 4 columns. So something like this:
Spoiler:


I only see 3 ways to put the column 4 if the column 3 is at the very top or bottom. And there are only 2 ways to put the column 4 if the column 3 is sandwiched between column 1 and 2. Column 3 is on the top twice, on the bottom twice and being sandwiched twice. Hence I only see (3 x 4) + (2 x 2) = 16 ways for 4 columns. What are another 8 ways for 4 columns that I missed?
Absolutely right!
+1 for 4 column version cheers

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Post by alexander mahone Fri Apr 05, 2013 12:04 pm

Is it 16 ways for 4 columns? I'm sure there's the less naive way to solve that though. Solving for the 5, 6, ... n columns obviously hard to visualize.
I only notice that we need to separate the case when the last column on the very top/bottom and when the last column being sandwiched. Number of ways that can be generated from each point when the last column on the very top/bottom is 2 for 2 columns, 3 for 3 and 4 columns, 4 for 5 and 6 columns, 5 for 7 and 8 columns, etc.
Then from each point when the last column being sandwiched eventually we also need to separate the case when the last column facing the edge between 2 columns and when it's not, because there's only 2 ways that can be generated when it's facing the edge between 2 columns and could be more when it's not Mad
Also, among the ways coming from each point where the last column on the very top/bottom or when it's being sandwiched but not facing the edge between 2 columns, 2 of them must end up on the very top/bottom and the rest will be being sandwiched.
And from 2 ways that can be generated from each point when the last column being sandwiched and facing the edge between 2 columns, 1 of them will face the edge between 2 columns and the other one won't.

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Post by Ion Creanga Mon Apr 22, 2013 7:49 pm

have a nice exercise for math fans that i just spent a lot of time yesterday in solving it..
polynom f = (x+i)^100 + (x-i)^100 , where i=sqrt(-1).
you have to proof that if f(x)=0 then x is a real number, not complex


(don't know if my english is clear enough for you to understand the problem)


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Post by Babun Wed Apr 24, 2013 10:29 am

tuddor wrote:have a nice exercise for math fans that i just spent a lot of time yesterday in solving it..
polynom f = , where i=sqrt(-1).
you have to proof that if f(x)=0 then x is a real number, not complex


(don't know if my english is clear enough for you to understand the problem)


That's just some back and forth playing with the index. You could write (x+y)^n as this:
Maths- Gym for Brainiacs  - Page 27 B043db415a899a4074efb54469ccf531

I'll write sum(k=0, n) for the big E-like sum sign and nCk for the binomial coefficient:

(x+i)^100 + (x-i)^100= sum(k=0,100)(100Ck)*x^(100-k)*i^k + sum(k=0,100)(100Ck)*x^(100-k)*(-i)^k=
---------------------------------
(-i)^k is the same as (-1)^k*i^k
----------------------------------
sum(k=0,100)(100Ck)*x^(100-k)*i^k + sum(k=0,100)(100Ck)*x^(100-k)*(-1)^k*i^k=

sum(k=0,100)(100Ck)*x^(100-k)*i^k *(1+(-1)^k)

---------------------------------------------------------------------------
1.for k=2n+1 1+(-1)^(2n+1)=0, for n element from natural numbers |N
2.for k=2n 1+(-1)^2n=2

So for all odd k the summand equals to zero hence I could leave them out.
----------------------------------------------------------------------------
What remains is this:
2*sum(n=0,100)(100C2n)*x^(100-2n)*i^2n

I'll do the rest later on ( lectures), basically i goes in 4 cycle (i,-1,-i,1, repeat,...). If someone is interested they could finish the proof.
P.S. one could argue i^2n could only be equal to 1 or -1 that's why the sum is made out of real numbers wtihout calculating much. The rest would follow from that statement.
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Post by Ion Creanga Thu Apr 25, 2013 10:02 pm

Babun, that's what i tried first too, used Newton's Bynom and i came too nothing, long calculations, plus it's hard to reach the conclusion this way..

There are two easier methods, one that i especially like.. you make this:

(x+i)^100=-(x-i)^100

now comes the beautiful part, you can use the module ..

module (x+i)^100=module (x-i)^100

so, module (x+i)=module (x-i)

you make the notation of a complex number x= a+bi, where a and b are real numbers

module a+(b+1)*i= module a+(b-1)*i
now we can can calculate the two modules of complex numbers..

sqrt(a^2+(b+1)^2)=sqrt(a^2+(b-1)^2)
a^2+b^2+2b+1=a^2+b^2-2b+1
, so after calculations
4*b=0, so b=0
if we replace b with 0 in x=a+bi, it makes x=a , where a is real number, so the conclusion is demonstrated
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Post by Babun Wed May 08, 2013 6:41 am

Yeah, your way is faster. I think we could also try to manipulate both terms with powers of (z²)^50 to get the is out of the way faster once you bring one of the left sided term to teh right with substraction.
alexander mahone wrote:Is it 16 ways for 4 columns? I'm sure there's the less naive way to solve that though. Solving for the 5, 6, ... n columns obviously hard to visualize.
I only notice that we need to separate the case when the last column on the very top/bottom and when the last column being sandwiched. Number of ways that can be generated from each point when the last column on the very top/bottom is 2 for 2 columns, 3 for 3 and 4 columns, 4 for 5 and 6 columns, 5 for 7 and 8 columns, etc.
Then from each point when the last column being sandwiched eventually we also need to separate the case when the last column facing the edge between 2 columns and when it's not, because there's only 2 ways that can be generated when it's facing the edge between 2 columns and could be more when it's not Mad
Also, among the ways coming from each point where the last column on the very top/bottom or when it's being sandwiched but not facing the edge between 2 columns, 2 of them must end up on the very top/bottom and the rest will be being sandwiched.
And from 2 ways that can be generated from each point when the last column being sandwiched and facing the edge between 2 columns, 1 of them will face the edge between 2 columns and the other one won't.

This problem is more than 100 years old. I wasn't trolling anyone but no one has found a connotation for all n element from natural numbers eco smile
You did your best get out of the way the first 4-6 eco smile
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Post by Raptorgunner Mon Aug 26, 2013 1:31 am

How do you make 1, 3, 4, and 6 equal 24 in a single digit consecutive equation?

I know its easy for you guys but I have been trying for 2 days and still don't get. Not really a math guy, not my subject.
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Post by Babun Mon Aug 26, 2013 11:37 pm

Raptorgunner wrote:How do you make 1, 3, 4, and 6 equal 24 in a single digit consecutive equation?

I know its easy for you guys but I have been trying for 2 days and still don't get. Not really a math guy, not my subject.
24=3*8=4*6=2*2*2*3
In the end, you need 4*6 or 3*8 somehow. 6 is already given. Try out some fraction, maybe you'll have more luck with that :coffee:
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Post by spanky Tue Aug 27, 2013 4:00 am

î though about doing it this way 1^-1 + 3^0 + 6^1+4^2
1^-1=1
3^0=1
6^1=6
4^2=16
1+1+6+16=24

or u could do it like this which is actually what i found when googling: "single digit consecutive equation"

1^3*6*4=24

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Post by Babun Tue Aug 27, 2013 1:11 pm

6/(1-3/4)=24
spanky wrote:î though about doing it this way 1^-1 + 3^0 + 6^1+4^2
1^-1=1
3^0=1
6^1=6
4^2=16
1+1+6+16=24

or u could do it like this which is actually what i found when googling: "single digit consecutive equation"

1^3*6*4=24

I don't think it's correct. 1^3 means 1*1*1 and you can use 1 once. Same story with the other examples, you use numbers that aren't given.
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Post by Raptorgunner Tue Aug 27, 2013 2:04 pm

Thanks guys, starting to get it now.
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Post by RealGunner Thu Sep 25, 2014 8:03 pm

Anyone still out there

2^2014 - 2^2013

?
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Post by spanky Thu Sep 25, 2014 9:46 pm

RealGunner wrote:Anyone still out there

2^2014 - 2^2013

?


2^2014=2^2013*2
2^2013*2-2^2013= 2^2013*(2-1)=2^2013

but looks more like infinity to me.
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Post by RealGunner Thu Sep 25, 2014 10:27 pm

You are alive :bow:

I saw this question elsewhere and the answers were split into either 2 or 2^2013 so I was confused

Normally I thought you would just take away the powers from each other?

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Post by spanky Thu Sep 25, 2014 10:35 pm

RealGunner wrote:You are alive :bow:

I saw this question elsewhere and the answers were split into either 2 or 2^2013 so I was confused

Normally I thought you would just take away the powers from each other?



yea im alive just inactive because football is at 1 am over here and sundays are working days. plus tv channels dont play european leagues, you have to subscribe the extra channels which are really expensive. anyways i still lurk around  Laughing  

regarding the problem you can also do it by logarithms
log2(x)=2014
log2(y)=2013  -subtract equations

log2(x)-log2(y)=2014-2013
log2(x/y)=1
x/y=2

you can take away the powers if it is multiplication or divison, with subtracting and adding it doesnt work like that unfortunately Sad
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Post by RealGunner Fri Sep 26, 2014 2:08 pm

Oh ffs forgot that :facepalm:

Yea that makes a lot more sense. Thanks
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Post by DuringTheWar Mon Mar 23, 2015 8:36 pm

Hello, is there anybody out there? I see this thread has been dead for a while, but ill ask this here anyway.

I have a question about dividing decimal numbers, something that's been driving me nuts for the last hour. Ok, the basic principles of division are clear, you want to find what goes into a number a given amount of times, ie 20/5 is asking what goes into 20 five times. Whole numbers are easy, even dividing a whole number by a part number is easy, you just have to think about it a different way, ie 50/0.1 just think how many times 0.1 goes into 50, so logically in that case the quotient is bigger than the dividend.

But I don't know how to reconcile this principle to dividing one part number by another. For example, 0.001/0.02, my calculator will tell me the answer is 0.05. What I don't understand is how 2 hundredths goes into 1 thousandth 5 hundredth times.

I can understand 1 tenth going into 50 five hundred times, because there the divisor is smaller than the dividend. But in this case the divident, 1000th, is smaller than the divisor, 100th, so how is it 100th goes into 1000th 5 times?
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Post by El Gunner Mon Mar 23, 2015 11:36 pm

loooooool didn't even understand what you just said.
Math is such an ass. Will probably fail my uni courses big time if I have to do it there. Laughing
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Post by Babun Tue Mar 24, 2015 3:52 pm

DuringTheWar wrote:Hello, is there anybody out there? I see this thread has been dead for a while, but ill ask this here anyway.

I have a question about dividing decimal numbers, something that's been driving me nuts for the last hour. Ok, the basic principles of division are clear, you want to find what goes into a number a given amount of times, ie 20/5 is asking what goes into 20 five times. Whole numbers are easy, even dividing a whole number by a part number is easy, you just have to think about it a different way, ie 50/0.1 just think how many times 0.1 goes into 50, so logically in that case the quotient is bigger than the dividend.

But I don't know how to reconcile this principle to dividing one part number by another. For example, 0.001/0.02, my calculator will tell me the answer is 0.05. What I don't understand is how 2 hundredths goes into 1 thousandth 5 hundredth times.

I can understand 1 tenth going into 50 five hundred times, because there the divisor is smaller than the dividend. But in this case the divident, 1000th, is smaller than the divisor, 100th, so how is it 100th goes into 1000th 5 times?

Read this first, son:
http://en.wikipedia.org/wiki/Scientific_notation


then this
Maths- Gym for Brainiacs  - Page 27 Exponent_Rules

0.001/0.02 = 10^(-3)/(2*10^(-2))=1/2 * 10^(-3+2)= 1/2 * 10^(-1) = 1/2*1/10 = 1/20 = 0.05

The easier way is to multiply with the highest 10 exponent. In case of 0.001/0.02 * 1000/1000 = 1/20
1000/1000 = 1 so by multiplying you don't change the result => allowed.
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Post by DuringTheWar Tue Mar 31, 2015 8:40 am

Babun got another question. Lets say you have two expressions, 2x + 5, and -1/2 + 3x. You are asked to add the expressions. So you start by combining like terms, 2x + 3x = 5x. But what do you do with the fraction that has a minus sign in front of it? Do you just subtract positive 1/2 from 5 as that is the same as adding negative 1/2 to 5? How would you write it?
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Post by elitedam Tue Mar 31, 2015 9:16 am

5 + (-1/2) = 5 - 1/2 = 4 1/2

(2x + 5) + (-1/2 + 3x) = 5x + 4 1/2
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Post by RealGunner Tue Apr 14, 2015 11:32 pm

http://ichef.bbci.co.uk/news/660/media/images/82299000/jpg/_82299333_24e94c5d-dc7b-4590-8f46-29d0da74c8b8.jpg

Babun, RO, Spanky pls respond
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Post by Cruijf Wed Apr 15, 2015 1:20 am

I love that one.

Spoiler:
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