Question

finding 100th digit after the decimal place of
( 2 + sqrt5)^500

Answer 1

I do have the solution but I do not understand it. In case anyone would like to see the solution ?

Answer 2

Hmm, I'm guessing maybe we need to use the binomial theorem... all the even powers on the sqrt 5 would give us whole numbers, all the odd powers would give multiples of sqrt 5.

Answer 3

I would like to see the solution, when you are online, post it, please.

Answer 4

http://yourmathguru.com/forums/topic/the-100th-digit

Answer 5

@mukushla look here if you can help then..

Answer 6

Or see
http://www.wolframalpha.com/input/?i=what+is+the+101+digit+of++%282+%2B+sqrt5%29%5E500+

Answer 7

are you sure wolfram is telling 100th digit after the decimal ?

Answer 8

is this a right representation ?
http://www.wolframalpha.com/input/?i=%28+greatest+integer+%5B+%2810%5E100+%29%282+%2B+sqrt5%29%5E500+%5D+%29+mod10

Answer 9

well yes but what if you want to know 3rd digit after decimal of 186.3349 ??
input will vary

Answer 10

yes, wolfram is a tool. Maybe in a few more years it will be a mathematician ...

Answer 11

well what I wanted to say was your input in wolfram was wrong

Answer 12

oh, wait. It is the 10^313 power, so that is not what you want.

Answer 13

so here is what you want

sage: N(( 2 + sqrt(5))^500,digits=420)
3.03012381665493808617352705454714235391879140029371493970855922136019868971187806629576822563956581117434290608358832071471866334615305238079729662622640319590069101303430750653363699328946679390632157734812572428455630808247147382467591044592551304368531540021043660638089720375493820280432347278412694880050000019999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999998e313

(the last 8 may not be accurate)

so it is 9.

Answer 14

I was just curious if we could get the answer using computation rather than analysis.

Answer 15

hmm.. well since we are confirmed answer is 9 , did you understand their method ?

Answer 16

Here is what they did.

(2+ sqrt5)^500 + (2 - sqrt5)^500 is an integer
use the binomial theorem to expand and note that you get terms like
nCm 2^m 5^{(n-m)/2} + nCm (-1)^(n-m) 2^m 5^{(n-m)/2}
when n-m is odd, these cancel. when n-m is even you get terms
nCm 2^(m+1) 5^{(n-m)/2}
which are integers.

next, we have 2-sqrt5 is about -0.236.....
raised to the 500th power (2-sqrt5)^500 will be positive and much less than 1
0 < (2-sqrt5)^500 << 1

we can take the log base 10 of (2-sqrt5)^500 to get an estimate of big (small ) this number is.
500*log(0.236...) (we use a positive base to avoid imaginary numbers)
we get about -313.5
in other words,
(2-sqrt5)^500 about = 10^-313.5
it is 0.0000... lots of zeros ...00000...digits

on the other hand, (2+sqrt5)^500 is some big number
nnnn.dddddd with an integer part and a fractional part
when we add (2-sqrt5)^500 to it, we get an integer (no decimals)

nnnnn...nnn.ddddddddddddddddddddddd
+000000000.000000000000000000eeeeee
___________________________________________
mmmmmm.000000000000000000000000

(btw, d is not necessarily the same digit here)
we know the eee part must add with the ddd part to give zeros, which means we are generating a carry of 1
at the 313 digit in we have
d + 0 + carry= 0
or because we know we have a carry of 1
d+1=0 plus carry
d must be 9, which gives 0 plus a carry.
Following that logic, all the d's in the 312 position or less must be 9, to generate 0's
therefore the 100th digit must be a 9

Answer 17

still not clear ?

Answer 18

I understood the log thing . but it bothers me that in this question you'll have to make use of calculator