Let N be the positive integer with 2018 decimal digits, all of them 1: that is N = 11111cdots111. What is the thousand digit after the decimal point of sqrt(N)?

1 Answer
Jun 29, 2017

3

Explanation:

Note that the given integer is 1/9(10^2018-1), so it has positive square root very close to 1/3(10^1009)

Note that:

(10^1009-10^-1009)^2 = 10^2018-2+10^-2018 < 10^2018-1

(10^1009-10^-1010)^2 = 10^2018-2/10+10^-2020 > 10^2018-1

So:

10^1009-10^-1009 < sqrt(10^2018-1) < 10^1009-10^-1010

and:

1/3(10^1009-10^-1009) < sqrt(1/9(10^2018-1)) < 1/3(10^1009-10^-1010)

The left hand side of this inequality is:

overbrace(333... 3)^"1009 times".overbrace(333...3)^"1009 times"

and the right hand side is:

overbrace(333... 3)^"1009 times".overbrace(333...3)^"1010 times"

So we can see that the 1000th decimal place is 3.