Given two integers K and N, the task is to find the count of integers from the range [2, N – 1] whose sum of prime divisors is K
Input: N = 20, K = 7
7 and 10 are the only valid numbers.
sumPFactors(7) = 7
sumPFactors(10) = 2 + 5 = 7
Input: N = 25, K = 5
Approach: Create an array sumPF where sumPF[i] stores the sum of prime divisors of i which can be easily calculated using the approach used in this article. Now, initialise a variable count = 0 and run a loop from 2 to N – 1 and for every element i if sumPF[i] = K then increment the count.
Below is the implementation of the above approach:
- Numbers in range [L, R] such that the count of their divisors is both even and prime
- Sum of numbers in a range [L, R] whose count of divisors is prime
- Sum of all prime divisors of all the numbers in range L-R
- Count the numbers < N which have equal number of divisors as K
- Program to find count of numbers having odd number of divisors in given range
- Find count of Almost Prime numbers from 1 to N
- Count numbers from range whose prime factors are only 2 and 3
- Count common prime factors of two numbers
- Count all the numbers less than 10^6 whose minimum prime factor is N
- Count numbers in a range having GCD of powers of prime factors equal to 1
- Queries for the difference between the count of composite and prime numbers in a given range
- Count Numbers in Range with difference between Sum of digits at even and odd positions as Prime
- Count of distinct sums that can be obtained by adding prime numbers from given arrays
- Absolute difference between the Product of Non-Prime numbers and Prime numbers of an Array
- Sum of all the prime divisors of a number
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