Maths Gym for Brainiacs
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Ion Creanga
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Re: Maths Gym for Brainiacs
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.
You didn't think the question would be this easy? Am I known for easy questions?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
Absolutely right!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?
+1 for 4 column version
Babun Fan Favorite
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Re: Maths Gym for Brainiacs
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
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.
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
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.
alexander mahone Hot Prospect
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Re: Maths Gym for Brainiacs
have a nice exercise for math fans that i just spent a lot of time yesterday in solving it..
polynom f = (x+i)^100 + (xi)^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)
polynom f = (x+i)^100 + (xi)^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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Re: Maths Gym for Brainiacs
That's just some back and forth playing with the index. You could write (x+y)^n as this: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)
I'll write sum(k=0, n) for the big Elike sum sign and nCk for the binomial coefficient:
(x+i)^100 + (xi)^100= sum(k=0,100)(100Ck)*x^(100k)*i^k + sum(k=0,100)(100Ck)*x^(100k)*(i)^k=

(i)^k is the same as (1)^k*i^k

sum(k=0,100)(100Ck)*x^(100k)*i^k + sum(k=0,100)(100Ck)*x^(100k)*(1)^k*i^k=
sum(k=0,100)(100Ck)*x^(100k)*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^(1002n)*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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Re: Maths Gym for Brainiacs
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=(xi)^100
now comes the beautiful part, you can use the module ..
module (x+i)^100=module (xi)^100
so, module (x+i)=module (xi)
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+(b1)*i
now we can can calculate the two modules of complex numbers..
sqrt(a^2+(b+1)^2)=sqrt(a^2+(b1)^2)
a^2+b^2+2b+1=a^2+b^22b+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
There are two easier methods, one that i especially like.. you make this:
(x+i)^100=(xi)^100
now comes the beautiful part, you can use the module ..
module (x+i)^100=module (xi)^100
so, module (x+i)=module (xi)
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+(b1)*i
now we can can calculate the two modules of complex numbers..
sqrt(a^2+(b+1)^2)=sqrt(a^2+(b1)^2)
a^2+b^2+2b+1=a^2+b^22b+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
Ion Creanga First Team
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Re: Maths Gym for Brainiacs
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.
You did your best get out of the way the first 46
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 numbersalexander 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
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.
You did your best get out of the way the first 46
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Re: Maths Gym for Brainiacs
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.
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.
Raptorgunner World Class Contributor
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Re: Maths Gym for Brainiacs
24=3*8=4*6=2*2*2*3Raptorgunner 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.
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
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Re: Maths Gym for Brainiacs
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
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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Re: Maths Gym for Brainiacs
6/(13/4)=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.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
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Re: Maths Gym for Brainiacs
Thanks guys, starting to get it now.
Raptorgunner World Class Contributor
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Re: Maths Gym for Brainiacs
Anyone still out there
2^2014  2^2013
?
2^2014  2^2013
?
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Re: Maths Gym for Brainiacs
RealGunner wrote:Anyone still out there
2^2014  2^2013
?
2^2014=2^2013*2
2^2013*22^2013= 2^2013*(21)=2^2013
but looks more like infinity to me.
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Re: Maths Gym for Brainiacs
You are alive
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?
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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Re: Maths Gym for Brainiacs
RealGunner wrote:You are alive
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
regarding the problem you can also do it by logarithms
log2(x)=2014
log2(y)=2013 subtract equations
log2(x)log2(y)=20142013
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
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Re: Maths Gym for Brainiacs
Oh ffs forgot that
Yea that makes a lot more sense. Thanks
Yea that makes a lot more sense. Thanks
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Re: Maths Gym for Brainiacs
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?
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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Re: Maths Gym for Brainiacs
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.
Math is such an ass. Will probably fail my uni courses big time if I have to do it there.
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Re: Maths Gym for Brainiacs
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
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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Re: Maths Gym for Brainiacs
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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Re: Maths Gym for Brainiacs
5 + (1/2) = 5  1/2 = 4 1/2
(2x + 5) + (1/2 + 3x) = 5x + 4 1/2
(2x + 5) + (1/2 + 3x) = 5x + 4 1/2
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Re: Maths Gym for Brainiacs
http://ichef.bbci.co.uk/news/660/media/images/82299000/jpg/_82299333_24e94c5ddc7b45908f4629d0da74c8b8.jpg
Babun, RO, Spanky pls respond
Babun, RO, Spanky pls respond
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Re: Maths Gym for Brainiacs
I love that one.
 Spoiler:
 Albert knows Bernard can't know just from the day. This means the month he was told doesn't include a single date that is unique in the sense it doesn't show up in another month. This rules out May (because May 19th, 19 doesn't show up elsewhere) and June (because of June 18th, 18 doesn't show up elsewhere).
This leaves July and August. When Bernard hears that Albert is sure he doesn't know, he goes through the same process we went through and narrows it down to July and August. The date he was given must be unique within those two months because he then knows. This rules out July 14th and August 14th.
Once Albert finds out Bernard knows, he also knows. If he also knows, then the month he was given must be July (if he was told August then the day could've been either August 15th or 17th and Bernard wouldn't have been able to figure it out. July however only has one day that fits the criteria).
This means her birthday is July 16! Let me know if you have any questions lol just explaining that made my head hurt.
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