A **P-smooth number** or **P-friable number ** is an integer whose largest prime factor is less than or equal to P. Given N and P, we need to write a program to check whether it is P-friable or not.

Examples:

Input : N = 24 , P = 7 Output : YES Explanation : The prime divisors of 24 are 2 and 3 only. Hence its largest prime factor is 3 which is less than or equal to 7, it is P-friable. Input : N = 22 , P = 5 Output : NO Explanation : The prime divisors are 11 and 2, hence 11>5, so it is not a P-friable number.

The approach will be to prime factorize the number and store the maximum of all the prime factors. We first divide the number by 2 if it is divisible, then we iterate from 3 to Sqrt(n) to get the number of times a prime number divides a particular number which reduces every time by n/i and store the prime factor i if its divides N. We divide our number n (by prime factors) by its corresponding smallest prime factor till n becomes 1. And if at the end n > 2, it means its a prime number, so we store that as a prime factor as well. At the end the largest factor is compared with p to check if it is p-smooth number or not.

## CPP

// CPP program to check if a number is // a p-smooth number or not #include <iostream> #include<math.h> using namespace std; // function to check if number n // is a P-smooth number or not bool check(int n, int p) { int maximum = -1; // prime factorise it by 2 while (!(n % 2)) { // if the number is divisible by 2 maximum = max(maximum, 2); n = n/2; } // check for all the possible numbers // that can divide it for (int i = 3; i <= sqrt(n); i += 2) { // prime factorize it by i while (n % i == 0) { // stores the maximum if maximum // and i, if i divides the number maximum = max(maximum,i); n = n / i; } } // if n at the end is a prime number, // then it a divisor itself if (n > 2) maximum = max(maximum, n); return (maximum <= p); } // Driver program to test above function int main() { int n = 24, p = 7; if (check(n, p)) cout << "yes"; else cout << "no"; return 0; }

## Java

// Java program to check if a number is // a p-smooth number or not import java.lang.*; class GFG{ // function to check if number n // is a P-smooth number or not static boolean check(int n, int p) { int maximum = -1; // prime factorise it by 2 while ((n % 2) == 0) { // if the number is divisible by 2 maximum = Math.max(maximum, 2); n = n/2; } // check for all the possible numbers // that can divide it for (int i = 3; i <= Math.sqrt(n); i += 2) { // prime factorize it by i while (n % i == 0) { // stores the maximum if maximum // and i, if i divides the number maximum = Math.max(maximum,i); n = n / i; } } // if n at the end is a prime number, // then it a divisor itself if (n > 2) maximum = Math.max(maximum, n); return (maximum <= p); } // Driver program to test // above function public static void main(String[] args) { int n = 24, p = 7; if (check(n, p)) System.out.println("yes"); else System.out.println("no"); } } // This code is contributed by // Smitha Dinesh Semwal

## Python3

# Python 3 program to # check if a number is # a p-smooth number or not import math # function to check if number n # is a P-smooth number or not def check(n, p) : maximum = -1 # prime factorise it by 2 while (not(n % 2)): # if the number is divisible by 2 maximum = max(maximum, 2) n = int(n/2) # check for all the possible numbers # that can divide it for i in range(3,int(math.sqrt(n)), 2): # prime factorize it by i while (n % i == 0) : # stores the maximum if maximum # and i, if i divides the number maximum = max(maximum,i) n = int(n / i) # if n at the end is a prime number, # then it a divisor itself if (n > 2): maximum = max(maximum, n) return (maximum <= p) # Driver program to test above function n = 24 p = 7 if (check(n, p)): print( "yes" ) else: print( "no" ) # This code is contributed by # Smitha Dinesh Semwal

Output:

yes

**Reference**:

http://oeis.org/wiki/P-smooth_numbers

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