mathschallenge.net

Problem

How many digits does the number 2¹⁰⁰⁰ contain?

Solution

As 2¹⁰⁰⁰ is not a multiple of 10, it follows that,
10^m 2¹⁰⁰⁰ 10^{m + 1}, where 10^m contains m + 1 digits.

Solving 2¹⁰⁰⁰ = 10^k, where m k m + 1

k = log 2¹⁰⁰⁰ = 1000 log 2 301.02999... , so m = [1000 log 2] = 301.

Hence 2¹⁰⁰⁰ contains 302 digits.

What is the least value of n, such that 2ⁿ contains exactly one million digits?