Question

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)`?

Answer 1

`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 `1000`th decimal place is `3`.