Answer to Question #89325 in Combinatorics | Number Theory for Aravind

Question

How do I find the answer? let n=9+99+999+…+9999…9;where the last number consist of 1999 digits of 9.How many times will the digit 1 appear in n ?

Answer 1

Example:

Consider simpler case first 9+99+999 = (10-1) + (100-1) + (1000-1) = 1110 - 3 = 1107

Solution:

n = 9+99+999+...+9999...9 = (10-1) + (100-1) + (1000-1) +...+ (10000...0-1) =

10 + 100 + 1000 + 10000...0 - 1999 = {10000...0 consists 1999 digits of 0 }

1111...10 - 1999 = {1111...10 consists 1999 digits of 1 }

1111...100000 + 11110 - 1999 = {1111...100000 consists 1995 digits of 1 }

1111...100000 + 9111 {1111...100000 consists 1995 digits of 1 }

Number n consists of (1995 + 3) digits of 1.

Answer : Digit 1 appear in n 1998 times.