POTW #374: Finding the 1000th Digit of the Square Root of a Large Number

[anemone]

Here is this week's POTW:

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Let $A=1+10+10^2+\cdots+10^{1997}$. Determine the 1000th digit after the decimal point of the square root of $A$ in base 10.

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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to the following members for their correct solution!(Cool)

1. MegaMoh
2. kaliprasad

Sample solution from other:

Spoiler

The answer is the same as the unit digit of $10^{1000}\sqrt{A}$. We have

$10^{1000}\sqrt{A}=10^{1000}\sqrt{\dfrac{10^{1998}-1}{9}}=\dfrac{\sqrt{10^{3998}-10^{2000}}}{3}$

Since $(10^{1997}-1)^2<10^{3998}-10^{2000}<(10^{1999}-4)^2$, so it follows that

$10^{1000}\sqrt{A}$ is between $\dfrac{10^{1999}-7}{3}=33\cdots33$ and $\dfrac{10^{1999}-4}{3}=33\cdots32$

Therefore the answer is 1.