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re: Daily Challenge #231 - Perfect Powers VIEW POST

FULL DISCUSSION
 

Python solution,

import math

def isPP(number : int) -> tuple:
    if(number > 3):
        for i in range(2, int(math.sqrt(number))+1):
            power = round(math.log(number)/math.log(i), 4)
            if power.is_integer():
                return (i, int(power))

Output,

print(isPP(4)) # output -> (2, 2)
print(isPP(9)) # output -> (3, 2)
print(isPP(5)) # output -> None
print(isPP(4)) # output -> (2, 2)
print(isPP(8)) # output -> (2, 3)
print(isPP(14)) # output -> None
print(isPP(184528125)) # output -> (45, 5)
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