Let n be the decimal integer 11...1 (with 1998 digits). What is the 1000th digit after the decimal point of √n?

**Solution**

Answer: 1.

*Fairly easy.*

A little experimentation suggests that if n is the integer with 2m decimal digits, all of them 1, then √n has m 3s, followed by a decimal point, followed by m 3s, followed by a 1, followed by some 6s. If true, that would imply that the required digit was 1.

n = (10^{2m} - 1)/9. Let x = (10^{m} - 10^{-m})/3, then x has m 3s followed by a decimal point, followed by m 3s. So we have to show that (x + 10^{-m-1})^{2} < n and (x + 2.10^{-m-1})^{2} > n. That is a straightforward computation.

© John Scholes

jscholes@kalva.demon.co.uk

12 Dec 1998