Given two positive integers L and R, the task is to count the elements from the range [L, R] whose prime factors are only 2 and 3.
Input: L = 1, R = 10
2 = 2
3 = 3
4 = 2 * 2
6 = 2 * 3
8 = 2 * 2 * 2
9 = 3 * 3
Input: L = 100, R = 200
Approach: Start a loop from L to R and for every element num:
- While num is divisible by 2, divide it by 2.
- While num is divisible by 3, divide it by 3.
- If num = 1 then increment the count as num has only 2 and 3 as its prime factors.
Print the count in the end.
Below is the implementation of the above approach:
- Count numbers from range whose prime factors are only 2 and 3 using Arrays | Set 2
- Count numbers in a given range whose count of prime factors is a Prime Number
- Print all numbers whose set of prime factors is a subset of the set of the prime factors of X
- Count all prime numbers in a given range whose sum of digits is also prime
- Check if a number exists having exactly N factors and K prime factors
- Count numbers in a range having GCD of powers of prime factors equal to 1
- Find number of factors of N when location of its two factors whose product is N is given
- Maximum number of prime factors a number can have with exactly x factors
- Count of numbers between range having only non-zero digits whose sum of digits is N and number is divisible by M
- Count of nodes in a Binary Tree whose child is its prime factors
- Count of Nodes whose both immediate children are its prime factors
- Find the row whose product has maximum count of prime factors
- Count pairs from a given range whose sum is a Prime Number in that range
- Find and Count total factors of co-prime A or B in a given range 1 to N
- K-Primes (Numbers with k prime factors) in a range
- Sum of numbers in a range [L, R] whose count of divisors is prime
- Count numbers in range [L, R] whose sum of digits is a Prime Number
- Super Ugly Number (Number whose prime factors are in given set)
- Sum of element whose prime factors are present in array
- Find prime factors of Array elements whose sum of exponents is divisible by K
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