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Thread: Find the 2013-th

  1. #1
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    Find the 2013-th

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

    help me!!
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  3. #3
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    Re: Find the 2013-th

    Quote Originally Posted by leezangqe View Post
    Let $\displaystyle N=1+10+10^2+...+10^{4023}$. Find the 2013-th digit after the decimal comma of $\displaystyle \sqrt N$
    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,
    $\displaystyle N=\underbrace{1111....1111}_\text{4024 1's}$
    let $\displaystyle 2n=4024$
    now$\displaystyle 9N<10^{2n} \implies 9N^2<10^{2n}N \implies 3N<10^n \sqrt{N}$(we first multiply both sides by N and then take square roots of both sides)
    also $\displaystyle 10^{2n}=9N+1$
    therefore, $\displaystyle 9N+6>10^{2n} \implies 9N^2+6N>10^{2n}N \implies 9N^2+6N+1>10^{2n}N \implies (3N+1)^2>10^{2n}N \implies (3N+1)>10^{n} \sqrt{N}$
    therefore we have showed that $\displaystyle 3N<10^{n}\sqrt{N}<3N+1$

    As $\displaystyle 10^{n}\sqrt{N}$ is between $\displaystyle 3N$ and $\displaystyle 3N+1$, its value would be something like $\displaystyle 3N.\underbrace{*******}_\text{digits after decimal point }$.
    Notice $\displaystyle 10^{n}\sqrt{N}=\underbrace{333...33}_\text{2n number of 3's}.******* $
    dividing both sides by $\displaystyle 10^n$ brings the decimal point forward by n places so value of $\displaystyle \sqrt{N}$ is 2012 3's followed by 2012 3's after the decimal point.We have to find the $\displaystyle 2013$th digit after the decimal point, which happens to be the first digit after decimal point in $\displaystyle 3N.\underbrace{*******}_\text{digits after decimal point }=10^{n}*\sqrt{N}$.so
    $\displaystyle (\underbrace{333...333}_\text{4024 3's}.*****)^{2}=\underbrace{111...111}_\text{4024 3's} *10^{2n}=1111...1110000...000$(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:
    $\displaystyle (\underbrace{333....3}_\text{2n digits}.5)^2=\underbrace{111....11}_\text{2n digits}.\underbrace{222....22}_\text{2n digits}$ (which is larger than required)
    $\displaystyle (\underbrace{333....3}_\text{2n digits}.3)^2=\underbrace{111....11}_\text{2n digits}.\underbrace{088....88}_\text{2n digits}$ (which is larger than required)
    $\displaystyle (\underbrace{333....3}_\text{2n digits}.2)^2=\underbrace{111....11}_\text{2n digits}.\underbrace{022....22}_\text{2n digits}$ (which is larger than required)
    $\displaystyle (\underbrace{333....3}_\text{2n digits}.1)^2=\underbrace{111....10}_\text{2n digits}.\underbrace{955...55}_\text{2n digits}$ (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
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