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Problem Of The Week #374 Jul 9th, 2019

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
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 POTW submission guidelines to find out how to submit your answers!
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
Congratulations to the following members for their correct solution!(Cool)

1. MegaMoh
2. kaliprasad

Sample solution from other:

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.
 
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