Given a number N, the task is to check whether the product of digits at even places of a number is divisible by K. If it is divisible, output “YES” otherwise output “NO”.
Examples:
Input: N = 5478, K = 5 Output: YES Since, 5 * 7 = 35, which is divisible by 5 Input: N = 19270, K = 2 Output: NO
Approach:
- Find product of digits at even places from right to left.
- Then check the divisibility by taking its modulo with ‘K’
- If modulo gives 0, output YES, otherwise output NO
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;// below function checks whether// product of digits at even places// is divisible by Kbool productDivisible(int n, int k){ int product = 1, position = 1; while (n > 0) { // if position is even if (position % 2 == 0) product *= n % 10; n = n / 10; position++; } if (product % k == 0) return true; return false;}// Driver codeint main(){ int n = 321922; int k = 3; if (productDivisible(n, k)) cout << "YES"; else cout << "NO"; return 0;} |
Java
// JAVA implementation of the above approach class GFG {// below function checks whether // product of digits at even places // is divisible by K static boolean productDivisible(int n, int k) { int product = 1, position = 1; while (n > 0) { // if position is even if (position % 2 == 0) { product *= n % 10; } n = n / 10; position++; } if (product % k == 0) { return true; } return false; }// Driver code public static void main(String[] args) { int n = 321922; int k = 3; if (productDivisible(n, k)) { System.out.println("YES"); } else { System.out.println("NO"); } }} |
Python3
# Python3 implementation of the # above approach # below function checks whether # product of digits at even places # is divisible by K def productDivisible(n, k): product = 1 position = 1 while n > 0: # if position is even if position % 2 == 0: product *= n % 10 n = n / 10 position += 1 if product % k == 0: return True return False# Driver coden = 321922k = 3if productDivisible(n, k) == True: print("YES")else: print("NO")# This code is contributed # by Shrikant13 |
C#
// C# implementation of the above approachusing System;class GFG{// below function checks whether// product of digits at even places// is divisible by Kstatic bool productDivisible(int n, int k){ int product = 1, position = 1; while (n > 0) { // if position is even if (position % 2 == 0) product *= n % 10; n = n / 10; position++; } if (product % k == 0) return true; return false;}// Driver codepublic static void Main(){ int n = 321922; int k = 3; if (productDivisible(n, k)) Console.WriteLine("YES"); else Console.WriteLine("NO");}}// This code is contributed // by Akanksha Rai(Abby_akku) |
PHP
<?php// PHP implementation of the // above approach// Below function checks whether// product of digits at even places// is divisible by Kfunction productDivisible($n, $k){ $product = 1; $position = 1; while ($n > 0) { // if position is even if ($position % 2 == 0) $product *= $n % 10; $n = (int)($n / 10); $position++; } if ($product % $k == 0) return true; return false;}// Driver code$n = 321922;$k = 3;if (productDivisible($n, $k)) echo "YES";else echo "NO";// This code is contributed by mits?> |
Javascript
<script>// Javascript implementation of the above approach// below function checks whether// product of digits at even places// is divisible by Kfunction productDivisible(n, k){ var product = 1, position = 1; while (n > 0) { // if position is even if (position % 2 == 0) product *= n % 10; n =parseInt(n / 10); position++; } if (product % k == 0) return true; return false;}// Driver codevar n = 321922;var k = 3;if (productDivisible(n, k)) document.write( "YES");else document.write( "NO");</script> |
Output:
YES
Time Complexity: O(log10n)
Auxiliary Space: O(1)
Method #2:Using string() method:
- Convert the integer to string then traverse the string and Multiply all even indices by storing it in the product.
- If the product is divisible by k then return True else False.
Below is the implementation:
C++
// C++ implementation of the// above approach#include <bits/stdc++.h>using namespace std;// Function checks whether// product of digits at even places// is divisible by Kbool productDivisible(int n, int k){ int product = 1; // Converting integer to string string num = to_string(n); // Traversing the string for (int i = 0; i < num.length(); i++) { if (i % 2 == 0) { product = product * (num[i] - '0'); } } if (product % k == 0) { return true; } else { return false; }}// Driver codeint main(){ int n = 321922; int k = 3; if (productDivisible(n, k)) { cout << "YES" << endl; } else { cout << "NO" << endl; }}// This code is contributed by phasing17 |
Java
// Java implementation of the// above approachimport java.util.*;class GFG { // Function checks whether // product of digits at even places // is divisible by K static boolean productDivisible(int n, int k) { int product = 1; // Converting integer to string String num = String.valueOf(n); // Traversing the string for (int i = 0; i < num.length(); i++) { if (i % 2 == 0) { product = product * (num.charAt(i) - '0'); } } if (product % k == 0) { return true; } else { return false; } } // Driver code public static void main(String[] args) { int n = 321922; int k = 3; if (productDivisible(n, k)) { System.out.println("YES"); } else { System.out.println("NO"); } }}// This code is contributed by phasing17 |
Python3
# Python3 implementation of the# above approach# Function checks whether# product of digits at even places# is divisible by Kdef productDivisible(n, k): product = 1 # Converting integer to string num = str(n) # Traversing the string for i in range(len(num)): if(i % 2 == 0): product = product*int(num[i]) if product % k == 0: return True return False# Driver coden = 321922k = 3if productDivisible(n, k) == True: print("YES")else: print("NO")# This code is contributed by vikkycirus |
C#
// C# implementation of the// above approachusing System;using System.Collections.Generic;class GFG{ // Function checks whether // product of digits at even places // is divisible by K static bool productDivisible(int n, int k) { int product = 1; // Converting integer to string string num = Convert.ToString(n); // Traversing the string for (int i = 0; i < num.Length; i++) { if (i % 2 == 0) { product = product * (num[i] - '0'); } } if (product % k == 0) { return true; } else { return false; } } // Driver code public static void Main(string[] args) { int n = 321922; int k = 3; if (productDivisible(n, k)) { Console.WriteLine("YES"); } else { Console.WriteLine("NO"); } }}// This code is contributed by phasing17 |
Javascript
<script>// JavaScript implementation of the// above approach // Function checks whether// product of digits at even places// is divisible by Kfunction productDivisible(n, k){ var product = 1 ; // Converting integer to string var num = n.toString() // Traversing the string for(let i = 0 ; i < num.length ; i++){ if(i % 2 == 0){ product = product * Number(num[i]) } } if (product % k == 0){ return true } else{ return false; }} // Driver codevar n = 321922var k = 3if(productDivisible(n, k)){ document.write("YES")}else{ document.write("NO")}</script> |
Output:
YES
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