Given N>0 and

N=PPPP.....PP(1998 digits)

what is the thousand digit after then decimal point of sqrt(N)????

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- Sep 1st 2010, 03:20 AMharshadN=PPPP.....PP(1998 digits)
Given N>0 and

N=PPPP.....PP(1998 digits)

what is the thousand digit after then decimal point of sqrt(N)???? - Sep 3rd 2010, 11:46 AMOpalg
Nobody else has attempted this, so I'll have a shot at it.

Start with the case where P = 9. Let N = 999...99 (2n digits). Eventually we'll want n = 999, but it's easier to start with a general n. Then $\displaystyle N = 10^{2n}-1$, and so

. . . . . $\displaystyle \begin{aligned}\sqrt N = 10^n(1-10^{-2n})^{1/2} &= 10^n(1-\tfrac1210^{-2n} - \tfrac1810^{-4n} - ...) \quad{\scriptstyle\text{(using the binomial expansion for $(1-x)^{1/2}$)}} \\ &= 10^n - 5\cdot10^{-n-1} - \tfrac1810^{-3n} - ... \\

&= \underbrace{999\cdots9}_{n\ 9\text{s}}.\underbrace{999\cdots9}_{n\ 9\text{s}}4999...\,. \end{aligned}$

So the (n+1)th digit after the decimal point is a 4. In particular, if n = 999 then the thousandth digit after the decimal point is a 4.

That deals with the case when P = 9. If P = 1, then you can divide $\displaystyle \sqrt N$ by 3 to see that the 1000th digit of $\displaystyle \sqrt{111...1}$ (1998 digits) is a 1. If P = 4 then the 1000th digit is a 3. But for other values of P, when $\displaystyle \sqrt P$ is irrational, I doubt whether there is any straightforward analytical way to answer the question.