Examine if given quantity has 7 divisors

Given a quantity N, the duty is to verify whether or not N has 7 divisors or not.

Examples:

Enter: 64
Output: 1 
Rationalization: 1, 2, 4, 8, 16, 32, 64 -> 7 divisors so output is 1

Enter: 100
Output: 0
Rationalization: 1, 2, 4, 5, 10, 20, 25, 50, 100 -> 8 divisors so output is 0

Enter: 729
Output: 1
Rationalization: 1, 2, 4, 8, 16, 32, 64 -> 7 divisors so output is 1

Method: The issue could be solved primarily based on the next mathematical statement:

The prime factorization of a quantity is:
N = p₁^e1 * p₂^e2 * . . . * p_n^en 
So variety of divisors = ( e₁ + 1 ) * (e₂ + 1) *. . . * (e_n + 1).

On this case, variety of divisors = 7 .
So, ( e₁ + 1 ) * (e₂ + 1) *. . . * (e_n + 1) = 7

7 is multiplication of at most 2 pure numbers. 
So, it may be solely written as ( e₁ + 1 ) * (e₂ + 1) = 1 * 7    
So, e₁ = 0  &  e₂ = 6  from above equation.

So, it’s clear that for 7 divisors just one case is feasible and that’s (prime quantity)⁶. Observe the steps to unravel the issue:

• Examine if N^(1/6) is a prime quantity or not.
• Whether it is prime quantity then output “Sure”.
• In any other case, the output can be “No”.

Under is the implementation of the above strategy: 

Time Complexity: O(logN)
Auxiliary Area: O(1)