Given a number N and numbers of digits in N, the task is to check whether the product of digits at even places of a number is divisible by sum of digits at odd place. If it is divisible, output “TRUE” otherwise output “FALSE”.
Examples:
Input: N = 2157
Output: TRUE
Since, 1 * 7 = 7, which is divisible by 2+5=7
Input: N = 1234
Output: TRUE
Since, 2 * 4 = 8, which is divisible by 1 + 3 = 4
Approach:
- Find product of digits at even places from right to left.
- Find sum of digits at odd places from right to left.
- Then check the divisibility of product by taking it’s modulo with sum
- If modulo gives 0, output TRUE, otherwise output FALSE
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 sum of digits at odd placesbool productSumDivisible(int n, int size){ int sum = 0, product = 1; while (n > 0) { // if size is even if (size % 2 == 0) { product *= n % 10; } // if size is odd else { sum += n % 10; } n = n / 10; size--; } if (product % sum == 0) return true; return false;}// Driver codeint main(){ int n = 1234; int len = 4; if (productSumDivisible(n, len)) cout << "TRUE"; else cout << "FALSE"; return 0;} |
Java
// JAVA implementation of the above approachclass GFG { // below function checks whether // product of digits at even places // is divisible by sum of digits at odd places static boolean productSumDivisible(int n, int size) { int sum = 0, product = 1; while (n > 0) { // if size is even if (size % 2 == 0) { product *= n % 10; } // if size is odd else { sum += n % 10; } n = n / 10; size--; } if (product % sum == 0) { return true; } return false; } // Driver code public static void main(String[] args) { int n = 1234; int len = 4; if (productSumDivisible(n, len)) { System.out.println("TRUE"); } else { System.out.println("FALSE"); } }} |
Python3
# Python 3 implementation of the above approach# Below function checks whether product# of digits at even places is divisible# by sum of digits at odd placesdef productSumDivisible(n, size): sum = 0 product = 1 while (n > 0) : # if size is even if (size % 2 == 0) : product *= n % 10 # if size is odd else : sum += n % 10 n = n // 10 size -= 1 if (product % sum == 0): return True return False# Driver codeif __name__ == "__main__": n = 1234 len = 4 if (productSumDivisible(n, len)): print("TRUE") else : print("FALSE")# This code is contributed by ChitraNayal |
C#
// C# implementation of the above approachusing System;class GFG { // below function checks whether // product of digits at even places // is divisible by K static bool productSumDivisible(int n, int size) { int sum = 0, product = 1; while (n > 0) { // if size is even if (size % 2 == 0) { product *= n % 10; } // if size is odd else { sum += n % 10; } n = n / 10; size--; } if (product % sum == 0) { return true; } return false; } // Driver code public static void Main() { int n = 1234; int len = 4; if (productSumDivisible(n, len)) Console.WriteLine("TRUE"); else Console.WriteLine("FALSE"); }} |
Javascript
<script>// Javascript implementation of the above approach// below function checks whether// product of digits at even places// is divisible by sum of digits at odd placesfunction productSumDivisible(n, size){ var sum = 0, product = 1; while (n > 0) { // if size is even if (size % 2 == 0) { product *= n % 10; } // if size is odd else { sum += n % 10; } n = parseInt(n / 10); size--; } if (product % sum == 0) return true; return false;}// Driver codevar n = 1234;var len = 4;if (productSumDivisible(n, len)) document.write("TRUE");else document.write("FALSE");// This code is contributed by noob2000.</script> |
PHP
<?php// PHP implementation of the above approach// Below function checks whether// product of digits at even // places is divisible by sum of// digits at odd placesfunction productSumDivisible($n, $size){ $sum = 0; $product = 1; while ($n > 0) { // if size is even if ($size % 2 == 0) { $product *= $n % 10; } // if size is odd else { $sum += $n % 10; } $n = $n / 10; $size--; } if ($product % $sum == 0) return true; return false;}// Driver code$n = 1234;$len = 4;if (productSumDivisible($n, $len)) echo "TRUE";else echo "FALSE";// This code is contributed by anuj_67..?> |
Output
TRUE
Time Complexity: O(logn)
Auxiliary Space: (1), since no extra space has been taken.
Method #2:Using string() method
Convert the integer to string then traverse the string and perform two operations
- Multiply all odd indices and store them in product
- Add all even indices and store them in sum
- If the product is divisible by sum then return True else False
Below is the implementation:
C++
// C++ implementation of the above approach#include <iostream>using namespace std;#include<string> // Below function checks whether product // of digits at even places is divisible // by sum of digits at odd placesbool productSumDivisible(int n) { int sum = 0; 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]); } else{ sum = sum+int(num[i]); } } if (product % sum == 0){ return true ; } else{ return false; } } // Driver code int main() { int n = 1234; if (productSumDivisible(n)) { cout<<"true"; } else{ cout<<"false"; } }// This code is contributed by akshitsaxena07 |
Java
// Java implementation of the above approachimport java.io.*;class GFG { // Below function checks whether product // of digits at even places is divisible // by sum of digits at odd places static boolean productSumDivisible(int n) { int sum = 0; 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*Character.getNumericValue(num.charAt(i)); } else{ sum = sum+Character.getNumericValue(num.charAt(i)); } } if (product % sum == 0){ return true ; } else{ return false; } } // Driver code public static void main (String[] args) { int n = 1234; if (productSumDivisible(n)) { System.out.println("TRUE"); } else{ System.out.println("FALSE"); } }}// This code is contributed by rag2127. |
Python3
# Python 3 implementation of the above approach# Below function checks whether product# of digits at even places is divisible# by sum of digits at odd placesdef productSumDivisible(n): sum = 0 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]) else: sum = sum+int(num[i]) if (product % sum == 0): return True return False# Driver codeif __name__ == "__main__": n = 1234 if (productSumDivisible(n)): print("TRUE") else: print("FALSE")# This code is contributed by vikkycirus |
C#
// C# implementation of the above approachusing System;class GFG{ // Below function checks whether product// of digits at even places is divisible// by sum of digits at odd placesstatic bool productSumDivisible(int n){ int sum = 0; int product = 1; // Converting integer to string string num = n.ToString(); // Traversing the string for(int i = 0; i < num.Length; i++ ) { if (i % 2 != 0) { product = product*(int)Char.GetNumericValue(num[i]); } else { sum = sum+(int)Char.GetNumericValue(num[i]); } } if (product % sum == 0) { return true; } else { return false; } }// Driver codestatic public void Main(){ int n = 1234; if (productSumDivisible(n)) { Console.WriteLine("TRUE"); } else { Console.WriteLine("FALSE"); }}}// This code is contributed by avanitrachhadiya2155 |
Javascript
<script>// JavaScript implementation of the above approach // Below function checks whether product// of digits at even places is divisible// by sum of digits at odd placesfunction productSumDivisible(n){ var sum = 0 var product = 1 // Converting integer to string var num = n.toString(); // Traversing the string for(i = 0 ; i < num.length ; i++ ) { if(i % 2 != 0){ product = product*Number(num[i]) } else{ sum = sum+Number(num[i]) } } if (product % sum == 0){ return true ; } else{ return false; }} // Driver code var n = 1234 if (productSumDivisible(n)){ document.write("TRUE") } else{ document.write("FALSE") } </script> |
Output
true
Time Complexity: O(d), where d is the number of digits in n
Auxiliary Space: (1), since no extra space has been taken.
Using List Comprehension:
Approach:
This approach uses list comprehension to extract the even and odd digits from the input number. It then computes the product of the even digits and the sum of the odd digits and checks if the product of even digits is divisible by the sum of odd digits.
Convert the input integer N into a list of its digits by converting it to a string, iterating over the string, and converting each character back to an integer using the int() function. This is done using the expression digits = [int(digit) for digit in str(N)].
Extract the even-indexed digits (i.e., the digits at positions 0, 2, 4, etc.) into a separate list even_digits using slicing. This is done using the expression even_digits = digits[::2].
Extract the odd-indexed digits (i.e., the digits at positions 1, 3, 5, etc.) into a separate list odd_digits using slicing. This is done using the expression odd_digits = digits[1::2].
Initialize a variable even_product to 1, which will hold the product of the even digits.
Compute the sum of the odd digits using the sum() function and store it in the variable odd_sum.
Iterate over the even digits using a for loop and multiply them together to compute their product, which is stored in the variable even_product. This is done using the code for digit in even_digits: even_product *= digit.
Test whether the even product is divisible by the odd sum using the modulo operator %. If the remainder is 0, then the even product is divisible by the odd sum and the function returns True. Otherwise, the even product is not divisible by the odd sum and the function returns False. This is done using the expression return even_product % odd_sum == 0.
C++
#include <algorithm>#include <iostream>#include <string>#include <vector>using namespace std;bool IsDivisible(int N){ // Convert N to a list of digits vector<int> digits; int tempN = N; while (tempN > 0) { digits.push_back(tempN % 10); tempN /= 10; } reverse(digits.begin(), digits.end()); // Separate even and odd positioned digits vector<int> evenDigits; vector<int> oddDigits; for (int i = 0; i < digits.size(); i++) { if ((i + 1) % 2 == 0) { evenDigits.push_back(digits[i]); } else { oddDigits.push_back(digits[i]); } } // Calculate the product of even digits and the sum of // odd digits int evenProduct = 1; int oddSum = 0; for (int digit : evenDigits) { evenProduct *= digit; } for (int digit : oddDigits) { oddSum += digit; } // Check if the even product is divisible by the odd sum return evenProduct % oddSum == 0;}int main(){ int N1 = 2157; int N2 = 1234; cout << boolalpha << IsDivisible(N1) << endl; // True cout << boolalpha << IsDivisible(N2) << endl; // True return 0;} |
Java
import java.util.ArrayList;import java.util.List;import java.util.Collections;public class DivisibilityChecker { public static boolean isDivisible(int N) { // Convert N to a list of digits List<Integer> digits = new ArrayList<>(); int tempN = N; while (tempN > 0) { digits.add(tempN % 10); tempN /= 10; } Collections.reverse(digits); // Separate even and odd positioned digits List<Integer> evenDigits = new ArrayList<>(); List<Integer> oddDigits = new ArrayList<>(); for (int i = 0; i < digits.size(); i++) { if ((i + 1) % 2 == 0) { evenDigits.add(digits.get(i)); } else { oddDigits.add(digits.get(i)); } } // Calculate the product of even digits and the sum of odd digits int evenProduct = 1; int oddSum = 0; for (int digit : evenDigits) { evenProduct *= digit; } for (int digit : oddDigits) { oddSum += digit; } // Check if the even product is divisible by the odd sum return evenProduct % oddSum == 0; } public static void main(String[] args) { int N1 = 2157; int N2 = 1234; System.out.println(isDivisible(N1)); // True System.out.println(isDivisible(N2)); // True }}// This code is contributed by akshitaguprzj3 |
Python3
def is_divisible(N): digits = [int(digit) for digit in str(N)] even_digits = digits[::2] odd_digits = digits[2::2] even_product = 1 odd_sum = sum(odd_digits) for digit in even_digits: even_product *= digit return even_product % odd_sum == 0N1 = 2157N2 = 1234print(is_divisible(N1)) # Trueprint(is_divisible(N2)) # True |
C#
using System;using System.Linq;class Program { static bool IsDivisible(int N) { // Convert N to a list of digits int[] digits = N.ToString() .ToCharArray() .Select(digit = > int.Parse(digit.ToString())) .ToArray(); // Separate even and odd positioned digits int[] evenDigits = digits .Where((digit, index) = > (index + 1) % 2 == 0) .ToArray(); int[] oddDigits = digits .Where((digit, index) = > (index + 1) % 2 != 0) .ToArray(); // Calculate the product of even digits and the sum // of odd digits int evenProduct = 1; int oddSum = oddDigits.Sum(); foreach(int digit in evenDigits) { evenProduct *= digit; } // Check if the even product is divisible by the odd // sum return evenProduct % oddSum == 0; } static void Main() { int N1 = 2157; int N2 = 1234; Console.WriteLine(IsDivisible(N1)); // True Console.WriteLine(IsDivisible(N2)); // True }} |
Javascript
function isDivisible(N) { // Convert N to a list of digits let digits = []; let tempN = N; while (tempN > 0) { digits.push(tempN % 10); tempN = Math.floor(tempN / 10); } digits.reverse(); // Separate even and odd positioned digits let evenDigits = []; let oddDigits = []; for (let i = 0; i < digits.length; i++) { if ((i + 1) % 2 === 0) { evenDigits.push(digits[i]); } else { oddDigits.push(digits[i]); } } // Calculate the product of even digits and the sum of odd digits let evenProduct = 1; let oddSum = 0; for (let digit of evenDigits) { evenProduct *= digit; } for (let digit of oddDigits) { oddSum += digit; } // Check if the even product is divisible by the odd sum return evenProduct % oddSum === 0;}let N1 = 2157;let N2 = 1234;console.log(isDivisible(N1)); // trueconsole.log(isDivisible(N2)); // true// This code is contributed by akshitaguprzj3 |
Output
True True
Time Complexity: O(n), where n is the number of digits in the input number.
Space Complexity: O(n)
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