Check if a number is divisible by all prime divisors of another number
Given two integers. We need to find if the first number x is divisible by all prime divisors of y.
Examples :
Input : x = 120, y = 75 Output : Yes Explanation : 120 = (2^3)*3*5 75 = 3*(5^2) 120 is divisible by both 3 and 5 which are the prime divisors of 75. Hence, answer is "Yes". Input : x = 15, y = 6 Output : No Explanation : 15 = 3*5. 6 = 2*3, 15 is not divisible by 2 which is a prime divisor of 6. Hence, answer is "No".
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A simple solution is to find all prime factors of y. For every prime factor, check if it divides x or not.
An efficient solution is based on below facts.
1) if y == 1, then it no prime divisors. Hence answer is “Yes”
2) We find GCD of x and y.
a) If GCD == 1, then clearly there are no common divisors of x and y, hence answer is “No”.
b) If GCD > 1, the GCD contains prime divisors which divide x also. Now, we have all unique prime divisor if and only if y/GCD has such unique prime divisor. So we have to find uniqueness for pair (x, y/GCD) using recursion.
C++
// CPP program to find if all prime factors// of y divide x.#include <bits/stdc++.h>using namespace std;// Returns true if all prime factors of y// divide x.bool isDivisible(int x, int y){ if (y == 1) return true; if (__gcd(x, y) == 1) return false; return isDivisible(x, y / gcd);}// Driver Codeint main(){ int x = 18, y = 12; if (isDivisible(x, y)) cout << "Yes" << endl; else cout << "No" << endl; return 0;} |
Java
// Java program to find if all// prime factors of y divide x.class Divisible{ public static int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); } // Returns true if all prime factors // of y divide x. static boolean isDivisible(int x, int y) { if (y == 1) return true; int z = gcd(x, y); if (z == 1) return false; return isDivisible(x, y / z); } // Driver program to test above functions public static void main(String[] args) { int x = 18, y = 12; if (isDivisible(x, y)) System.out.println("Yes"); else System.out.println("No"); }}// This code is contributed by Prerna Saini |
Python3
# python program to find if all# prime factors of y divide x.def gcd(a, b): if(b == 0): return a else: return gcd(b, a % b) # Returns true if all prime# factors of y divide x.def isDivisible(x,y): if (y == 1): return 1 z = gcd(x, y); if (z == 1): return false; return isDivisible(x, y / z);# Driver Codex = 18y = 12if (isDivisible(x, y)): print("Yes")else: print("No")# This code is contributed by Sam007 |
C#
// C# program to find if all// prime factors of y divide x.using System;class GFG { public static int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); } // Returns true if all prime factors // of y divide x. static bool isDivisible(int x, int y) { if (y == 1) return true; int z = gcd(x, y); if (z == 1) return false; return isDivisible(x, y / z); } // Driver program to test above functions public static void Main() { int x = 18, y = 12; if (isDivisible(x, y)) Console.WriteLine("Yes"); else Console.WriteLine("No"); }}// This code is contributed by vt_m. |
PHP
<?php// PHP program to find if all// prime factors of y divide x.function gcd ($a, $b){ return $b == 0 ? $a : gcd($b, $a % $b);}// Returns true if all prime// factors of y divide x.function isDivisible($x, $y){ if ($y == 1) return true; $z = gcd($x, $y); if ($z == 1) return false; return isDivisible($x, $y / $z);}// Driver Code$x = 18;$y = 12;if (isDivisible($x, $y)) echo "Yes";else echo "No";// This code is contributed by Sam007?> |
Javascript
<script>// Javascript program to find if all// prime factors of y divide x.function gcd(a , b) { return b == 0 ? a : gcd(b, a % b); }// Returns true if all prime factors// of y divide x.function isDivisible(x , y){ if (y == 1) return true; var z = gcd(x, y); if (z == 1) return false; return isDivisible(x, y / z);}// Driver program to test above functionsvar x = 18, y = 12;if (isDivisible(x, y)) document.write("Yes");else document.write("No");// This code is contributed by Amit Katiyar</script> |
Output :
Yes
Time Complexity:Time complexity for calculating GCD is O(log min(x, y)), and recursion will terminate after log y steps because we are reducing it by a factor greater than one. Overall Time complexity: O(log2y)
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