What is the 100th digit after the decimal point in the following expression:
If any one can help me regain my sanity I'll really appreciate it.
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joekz wrote:I've been working on this for 2 days, i don't feel like I'm getting any closer.
What is the 100th digit after the decimal point in the following expression:( \sqrt{5} + \sqrt {7} )^{2011}
If any one can help me regain my sanity I'll really appreciate it.
Macbi wrote:Hint:Spoiler:
joekz wrote:Xias thanks a lot! I'm at work right now but i will definitively try that path when ill get home.Macbi wrote:Hint:Spoiler:Spoiler:
addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk.
Yakk wrote:We cannot completely ignore the 1s part until the very end, because those factors end up infecting our non-ones parts when we multiply.
Macbi wrote:Hint:Spoiler:
Xias wrote:Yakk wrote:We cannot completely ignore the 1s part until the very end, because those factors end up infecting our non-ones parts when we multiply.
Right, but at the end, we'll have a 1s part and a non-1s part, and only the latter matters. It's for that reason though that actually calculating the non-1s part by hand would be difficult.Macbi wrote:Hint:Spoiler:
I'm curious now. For n=>even, you end up with double the integer part. What exactly happens when n=>odd, and how can we use this to our advantage?
sundalius wrote:Wait...
Wouldn't it just be:
(Radical 5+Radical 7) ^2
Then under distributive property:
5+7
Then to derive the answer:
5+7=12 then 12^2011
This was my first thought. Please explain how it's wrong or it'll bug me.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
Xias wrote:Yakk wrote:We cannot completely ignore the 1s part until the very end, because those factors end up infecting our non-ones parts when we multiply.
Right, but at the end, we'll have a 1s part and a non-1s part, and only the latter matters. It's for that reason though that actually calculating the non-1s part by hand would be difficult.Macbi wrote:Hint:Spoiler:
I'm curious now. For n=>even, you end up with double the integer part. What exactly happens when n=>odd, and how can we use this to our advantage?
addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk.
addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk.
sundalius wrote:Wait...
Wouldn't it just be:
(Radical 5+Radical 7) ^2
Then under distributive property:
5+7
Then to derive the answer:
5+7=12 then 12^2011
This was my first thought. Please explain how it's wrong or it'll bug me.
Macbi wrote:joekz wrote:Xias thanks a lot! I'm at work right now but i will definitively try that path when ill get home.Macbi wrote:Hint:Spoiler:Spoiler:
Ah, for some reason I thought we were in 2010. I now have no idea.
(Wolfram Alpha says the answer is 4.)
Kick wrote:Yes, W|A has the 100th digit as being 4, but is that not just the 100th digit and not the 100th digit after the decimal place?
EDIT: I am no mathematician, I just use W|A a lot...
t1mm01994 wrote:sundalius wrote:Wait...
Wouldn't it just be:
(Radical 5+Radical 7) ^2
Then under distributive property:
5+7
Then to derive the answer:
5+7=12 then 12^2011
This was my first thought. Please explain how it's wrong or it'll bug me.
http://en.wikipedia.org/wiki/Freshmans_dream
Quite relevant. Basically, following your train of thought
(2+3)^2 = 2^2 + 3^2
5^2 = 4+9
25 = 13 and that doesnt quite sound right to me.. A search for "expanding brackets" might help you as well.
mfb wrote:After looking at your pdf:
The expression inside the "[ ]" is ~10^1071, which is a lot (~300 orders of magnitude) smaller than the original number. I wonder where this comes from.
mfb wrote:Well, (\sqrt{7}+\sqrt{5})^2 = 12+2\sqrt{35}
Change \sqrt{5} to \sqrt{7}+\sqrt{5} in your b_{n} definition and you have the original problem.
So after your correction, the whole work looks a bit pointless.
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