Problem Description

The 5-digit number, 16807=7⁵, is also a fifth power. Similarly, the 9-digit number, 134217728=8⁹, is a ninth power.

How many n-digit positive integers exist which are also an nth power?

Analysis

Note: log() = log₁₀; ln() = log_e. One way to know how many digits a number has is to use int(log n) +1 or if you don’t have a log function then use the natural log function as int(ln n / ln 10) +1. For this problem n cannot be greater then 10 since 10ⁿ is always n+1 digit long.

So we need to create a function f such that f(n) = upper limit of n. Using our log function and using 9 as an example we define f(9) as int (1 / (1 – log₁₀ 9)) or 21 such that 9ⁿ < 10^n-1. Thus 9ⁿ is n digits long when n ≤ 21. We repeat this for 1 through 8 and sum for an answer.

Solution

Comments

We didn’t see a need to write a program for this one.