Let NN be the positive integer with 20182018 decimal digits, all of them 1: that is N = 11111cdots111N=11111⋯111. What is the thousand digit after the decimal point of sqrt(N)√N?

1 Answer
George C.
Jun 29, 2017

33

Explanation:

Note that the given integer is 1/9(10^2018-1)19(102018−1), so it has positive square root very close to 1/3(10^1009)13(101009)

Note that:

(10^1009-10^-1009)^2 = 10^2018-2+10^-2018 < 10^2018-1(101009−10−1009)2=102018−2+10−2018<102018−1

(10^1009-10^-1010)^2 = 10^2018-2/10+10^-2020 > 10^2018-1(101009−10−1010)2=102018−210+10−2020>102018−1

So:

10^1009-10^-1009 < sqrt(10^2018-1) < 10^1009-10^-1010101009−10−1009<√102018−1<101009−10−1010

and:

1/3(10^1009-10^-1009) < sqrt(1/9(10^2018-1)) < 1/3(10^1009-10^-1010)13(101009−10−1009)<√19(102018−1)<13(101009−10−1010)

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.

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