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:
- Count number of integers less than or equal to N which has exactly 9 divisors
- Check if a number is divisible by all prime divisors of another number
- 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 if a number can be expressed as a product of exactly K prime divisors
- Count of the non-prime divisors of a given number
- Numbers in range [L, R] such that the count of their divisors is both even and prime
- Maximum count of pairwise co-prime and common divisors of two given numbers
- Number which has the maximum number of distinct prime factors in the range M to N
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- Find the row whose product has maximum count of prime factors
- Count of Nodes which has Prime Digit sum weight in a Tree
- Generating all divisors of a number using its prime factorization
- Product of divisors of a number from a given list of its prime factors
- Count occurrences of a prime number in the prime factorization of every element from the given range
- Sum of all prime divisors of all the numbers in range L-R
- Check if the binary representation of a number has equal number of 0s and 1s in blocks
- Quick ways to check for Prime and find next Prime in Java
- Count all prime numbers in a given range whose sum of digits is also prime
- Check if a number exists with X divisors out of which Y are composite
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Improved By : AnkitRai01