Number of zeros after the decimal point

$10^{320}-\sqrt{10^{640}-1}= 10^{320}-\sqrt{10^{640}-1} \times \frac{10^{320}+\sqrt{10^{640}-1}}{10^{320}+\sqrt{10^{640}-1}}=\frac{1}{10^{320}+\sqrt{10^{640}-1}}$

Since $\sqrt{10^{640}-1}<10^{320}$ ,

$\frac{5}{10^{321}} = \frac{1}{2 \times 10^{320}} < \frac{1}{10^{320}+\sqrt{10^{640}-1}} < \frac{1}{10^{320}}$

$\frac{5}{10^{321}}$ has $320$ zeros, after decimal point. a number greater than it, should have less or equal number of zeros ( $320$ or less). $\frac{1}{10^{320}}$ has $319$ zeros, after decimal point. a number less than it, should have more number of zeros ( $320$ or more).

Consider in general $x_n=10^n - \sqrt{10^{2n}-1}$ . By binomial expansion,

We have $\begin{aligned} x_n &=10^n \left(1 - \sqrt{1-10^{-2n}} \right) \\ &=10^n \left(1-\left(1-\frac12 10^{-2n}+\cdots\right)\right) \\ &=10^n \left(\frac12 10^{-2n}+\cdots\right) \\ &=\frac12 10^{-n}+\cdots \\ &=5 \times 10^{-n-1}+\cdots \end{aligned}$

which has $n$ zeroes immediately after the decimal point.

$\sqrt {{10}^{640} - 1} \\ \approx \sqrt {(10^{320})^2 - 2 \times 10^{320} \times \dfrac 1{2 \times 10^{320}} + (\dfrac 1{2 \times 10^{320}})^2} \\ = \sqrt {(10^{320} - \dfrac 1{2 \times 10^{320}})^2} \\ = 10^{320} - \dfrac 1{2 \times 10^{320}} \\ \therefore 10^{320} - \sqrt {{10}^{640} - 1} \approx \dfrac 1{2 \times 10^{320}} = 0.5E(-320)$

Note that $10^{320} - \sqrt{10^{640}-1} = \dfrac {(10^{320} - \sqrt{10^{640}-1})(10^{320} + \sqrt{10^{640}-1})}{10^{320} + \sqrt{10^{640}-1}} = \dfrac 1{10^{320} + \sqrt{10^{640}-1}}$ and:

$\begin{aligned} \frac 1{10^{320}} < & \frac 1{10^{320} + \sqrt{10^{640}-1}} < \frac 1{2 \times 10^{320}} \\ 1 \times 10^{-320} < & 10^{320} - \sqrt{10^{640}-1} < 5 \times 10^{-321} \\ 0.\underbrace{00000 \cdots 00000}_{\text{Number of 0s}=319}1 < & 10^{320} - \sqrt{10^{640}-1} < 0.\underbrace{00000 \cdots 00000}_{\text{Number of 0s}=320}5 \end{aligned}$

The number of consecutive zeros after the decimal point of $10^{320} - \sqrt{10^{640}-1}$ is $\boxed{320}$ .

Consider the function $f(x)=\sqrt{x}$ Its derivative is $f'(x)=\frac{1}{2\sqrt{x}}$ We have $f(x+dx)=f(x)+f'(x)dx$ and to a very good approximation, we can set $x=10^{640}$ and replace $dx$ by $-1$ : $\sqrt{10^{640}-1}\approx \sqrt{10^{640}}-\frac{1}{2\sqrt{10^{640}}}=10^{320}-0.5×10^{-320}$ so that given expression is $10^{320}-\sqrt{10^{640}-1}\approx 0.5×10^{-320}$ which has 320 zeros between the decimal point and the 5.

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For those questioning the accuracy of this

This is in fact a 1st order Taylor expansion of $f(x-1)$ at $x={10^{640}}$ . We could add subsequent terms and see their contribution vanish very quickly:

$f(x-1)=f(x)-f'(x)+\frac{f''(x)}{2}-...+\frac{(-1)^nf^{(n)}(x)}{n!}+ ...$ using the derivatives

$f(x)=x^{0.5}, f'(x)=0.5x^{-0.5}, f''(x)=-0.25x^{-1.5}, f^{(3)}(x)=0.375x^{-2.5}$ etc

$10^{320}-f(x-1)=0.5×10^{-320}+0.125×10^{-960}+0.375/6×10^{-1600}...$

So after the 320 zeros comes '5', followed by 639 0's, followed by '125', followed by 638 0's, followed by '625', etc.