Let n be the decimal integer 11...1 (with 1998 digits). What is the 1000th digit after the decimal point of √n?
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 = (102m - 1)/9. Let x = (10m - 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
12 Dec 1998