# Check if a number is an Achilles number or not in Python

PythonServer Side ProgrammingProgramming

Suppose we have a number n; we have to check whether n is an Achilles number or not. As we know a number is Achilles number when a number that is powerful (A number N is called Powerful Number when for every prime factor p of it, p^2 also divides it) but not a perfect power. Some of the examples of Achilles numbers are: 72, 108, 200, 288, 392, 432, 500, 648, 675, 800, 864, 968, 972, 1125.

So, if the input is like 108, then the output will be True, as 6 and 36 both divide it and it is not perfect square.

To solve this, we will follow these steps −

• Define a function check_powerful() . This will take n
• while n mod 2 is same, do
• p := 0
• while n mod 2 is same as 0, do
• n := n / 2
• p := p + 1
• if p is same as 1, then
• return False
• p := integer of (square root of n) + 1
• for factor in range 3 to p, increase by 2, do
• p := 0
• while n mod factor is same as 0, do
• n := n / factor
• p := p + 1
• if p is same as 1, then
• return False
• return true when (n is same as 1)
• Define a function check_power() . This will take a
• if a is same as 1, then
• return True
• p := integer of (square root of n) + 1
• for i in range 2 to a, increase by 1, do
• val := log(a) / log(i) [all base e]
• if (val - integer part of (val)) < 0.00000001, then
• return True
• return False
• From the main method, do the following −
• if check_powerful(n) is same as True and check_power(n) is same as False, then
• return True
• otherwise,
• return False

## Example

Let us see the following implementation to get better understanding −

Live Demo

from math import sqrt, log
def check_powerful(n):
while (n % 2 == 0):
p = 0
while (n % 2 == 0):
n /= 2
p += 1
if (p == 1):
return False
p = int(sqrt(n)) + 1
for factor in range(3, p, 2):
p = 0
while (n % factor == 0):
n = n / factor
p += 1
if (p == 1):
return False
return (n == 1)
def check_power(a):
if (a == 1):
return True
p = int(sqrt(a)) + 1
for i in range(2, a, 1):
val = log(a) / log(i)
if ((val - int(val)) < 0.00000001):
return True
return False
def isAchilles(n):
if (check_powerful(n) == True and check_power(n) == False):
return True
else:
return False
n = 108
print(isAchilles(n))

## Input

108

## Output

True
Published on 27-Aug-2020 13:38:56