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## Project Euler 63: Count how many n-digit positive integers exist which are also an nth power

#### 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

One way to know how many digits a number has is to use the formula: int(log10 n) +1. For this problem n cannot be greater than 10 since 10n is always n+1 digits long.

So we need to create a function 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 < 10n-1. Thus 9n is n digits long when n ≤ 21. We repeat this same process for 1 through 8 and sum the results for an answer.

#### Project Euler 63 Solution

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