An infinite sequence of decimal digits is obtained by writing the positive integers in order: 123456789101112131415161718192021 ... . Define f(n) = m if the 10^{n} th digit forms part of an m-digit number. For example, f(1) = 2, because the 10th digit is part of 10, and f(2) = 2, because the 100th digit is part of 55. Find f(1987).

**Solution**

Answer: 1984.

*Easy.*

The first 9 numbers have 1 digit, the next 90 2 digits, the next 900 3 digits and so on. So the m-digit numbers and below constitute the first 9(1 + 2.10 + ... + m.10^{m-1}) = 9(1 - (m+1) 10^{m} + m 10^{m+1})/(1 - 10)^{2} = 1/9 (1 - (m+1) 10^{m} + m 10^{m+1}) digits. For m = 1983 this is about 0.2 10^{1987} and for m = 1984 about 2 10^{1987}. So the 10^{1987}th digit is in a 1984-digit number.

© John Scholes

jscholes@kalva.demon.co.uk

7 Oct 1999