Given an integer N, the task is to check if the count of divisors of N is prime or not.
Input: N = 13
The divisor count is 2 (1 and 13) which is prime.
Input: N = 8
The divisors are 1, 2, 4 and 8.
Approach: Please read this article to find the count of divisors of a number. So find the maximum value of i for every prime divisor p such that N % (pi) = 0. So the count of divisors gets multiplied by (i + 1). The count of divisors will be (i1 + 1) * (i2 + 1) * … * (ik + 1).
It can now be seen that there can only be one prime divisor for the maximum i and if N % pi = 0 then (i + 1) should be prime. The primality can be checked in sqrt(n) time and the prime factors can also be found in sqrt(n) time. So the overall time complexity will be O(sqrt(n)).
Below is the implementation of the above approach:
- Check if a number is divisible by all prime divisors of another number
- Numbers in range [L, R] such that the count of their divisors is both even and prime
- Generating all divisors of a number using its prime factorization
- Count occurrences of a prime number in the prime factorization of every element from the given range
- Count number of integers less than or equal to N which has exactly 9 divisors
- Find sum of divisors of all the divisors of a natural number
- Find sum of inverse of the divisors when sum of divisors and the number is given
- Check whether the given number is Wagstaff prime or not
- Sum of all prime divisors of all the numbers in range L-R
- Check if a number can be written as a sum of 'k' prime numbers
- Quick ways to check for Prime and find next Prime in Java
- Count Divisors of n in O(n^1/3)
- Count Divisors of Factorial
- Count divisors of array multiplication
- Find the largest good number in the divisors of given number N
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Improved By : AnkitRai01