#    
## Problem statement
How many $n$-digit positive integers exist that are also an $nth$ power?

```
ex => 16807 is a 5-digit number and also 7^5.
```

#    
## Solution approach

Any `n`-digit number `x` can be represented as follows:
```
10^(n-1) < x < 10^n 
```

We will run a `while` loop for `n` from 1, and increment the result by `largest`-`least` until `least` > 10.

1.  `least` is a number whose $nth$ power is the smallest `n`-digit $nth$ power of any number.

2.  `largest` is a number whose $nth$ power is the largest `n`-digit $nth$ power of any number.

### Example
- For `n`=2:
    - `least` = 4 because $3^2$ is 9, which is 1-digit.
    
    - `largest` = 9 because $10^2$ is 100, which is 3-digit.

    - So, we will increment the result by `largest`-`least`+1, i.e., 9-4+1=6.
     


from math import ceil
# ceil(x) returns least whole number greater than x

result=0
least=1
largest=9
n=1
while least<10 :
    least = ceil((10**(n-1))**(1/n))
    
    result += largest-least +1
    # +1 because range of (least,largest) is inclusive
    n+=1

print(result)


Project Euler 63: Powerful digit counts

Problem statement

Solution approach

Example