Putnam 1998 Problem B5 Let n be the decimal integer 11...1 (with 1998 digits). What is the 1000th digit after the decimal point of √n? Solution
: 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. That is a straightforward computation. Putnam 1998 © John Scholes jscholes@kalva.demon.co.uk 12 Dec 1998