Find the 2013-th

Quote:

Originally Posted by leezangqe

Let $N=1+10+10^2+...+10^{4023}$ . Find the 2013-th digit after the decimal comma of $\sqrt N$

If you try to the add the terms you will notice that you will get a long string of 1's.
so you will find,
$N=\underbrace{1111....1111}_\text{4024 1's}$
let $2n=4024$
now $9N<10^{2n} \implies 9N^2<10^{2n}N \implies 3N<10^n \sqrt{N}$ (we first multiply both sides by N and then take square roots of both sides)
also $10^{2n}=9N+1$
therefore, $9N+6>10^{2n} \implies 9N^2+6N>10^{2n}N \implies 9N^2+6N+1>10^{2n}N \implies (3N+1)^2>10^{2n}N \implies (3N+1)>10^{n} \sqrt{N}$
therefore we have showed that $3N<10^{n}\sqrt{N}<3N+1$

As $10^{n}\sqrt{N}$ is between $3N$ and $3N+1$ , its value would be something like $3N.\underbrace{*******}_\text{digits after decimal point }$ .
Notice $10^{n}\sqrt{N}=\underbrace{333...33}_\text{2n number of 3's}.*******$
dividing both sides by $10^n$ brings the decimal point forward by n places so value of $\sqrt{N}$ is 2012 3's followed by 2012 3's after the decimal point.We have to find the $2013$ th digit after the decimal point, which happens to be the first digit after decimal point in $3N.\underbrace{*******}_\text{digits after decimal point }=10^{n}*\sqrt{N}$ .so
$(\underbrace{333...333}_\text{4024 3's}.*****)^{2}=\underbrace{111...111}_\text{4024 3's} *10^{2n}=1111...1110000...000$ (with 2n 1's and 2n 0's)
Noe in order to find the first digit after decimal point, we use approximation.we use just one decimal place and square it to make it just small(or the first small value) below our intended square number.
After working with small numbers, a pattern emerge:
$(\underbrace{333....3}_\text{2n digits}.5)^2=\underbrace{111....11}_\text{2n digits}.\underbrace{222....22}_\text{2n digits}$ (which is larger than required)
$(\underbrace{333....3}_\text{2n digits}.3)^2=\underbrace{111....11}_\text{2n digits}.\underbrace{088....88}_\text{2n digits}$ (which is larger than required)
$(\underbrace{333....3}_\text{2n digits}.2)^2=\underbrace{111....11}_\text{2n digits}.\underbrace{022....22}_\text{2n digits}$ (which is larger than required)
$(\underbrace{333....3}_\text{2n digits}.1)^2=\underbrace{111....10}_\text{2n digits}.\underbrace{955...55}_\text{2n digits}$ (which is just smaller as 111...11>111...10)
So the required first digit after decimal point should be 1
or the 2013th digit should be 1.
Hope I explained clearly.Explaining this was hard (Whew)