Question

how to find the decimal expansion for 2 power 5 × 5 powe2

Answer 1

Prove that there is a power of 2 whose decimal representation starts with the digits 1999. ... I did some algebraic manipulations, such as: A = 2^m - 1999*10^n = 2^n*(2^(m-n) - 1999*5^n) and deduced that: 2^n || A ==> A = 2^n * b, where b is odd. Also, 1 <= A <= 10^n - 1 ==> 1 <= b <= 5^n - 1. And that was ...