Divisibility by 12 for a large number
Given a large Number the task is to check the number is divisible by 12 or not .
Examples :
Input : 12244824607284961224 Output : Yes Input : 92387493287593874594898678979792 Output : No
This is very simple approach. if a number is divisible by 4 and 3 then number is divisible by 12.
Point 1. If the last two digit of the number is divisible by 4 then number is divisible by 4. Please see divisibility by 4 for large numbers for details.
Point 2. if sum of all digit of a number divide by 3 then number is divisible by 3. Please see divisibility by 3 for large numbers for details.
CPP
// C++ Program to check if // number is divisible by 12 #include <iostream> using namespace std; bool isDvisibleBy12(string num) { // if number greater then 3 if (num.length() >= 3) { // fiind last digit int d1 = (int)num[num.length() - 1]; // no is odd if (d1 % 2 != 0) return (0); // find second last digit int d2 = (int)num[num.length() - 2]; // find sum of all digits int sum = 0; for (int i = 0; i < num.length(); i++) sum += num[i]; return (sum % 3 == 0 && (d2 * 10 + d1) % 4 == 0); } else { // if number is less then // or equal to 100 int number = stoi(num); return (number % 12 == 0); } } // Driver function int main() { string num = "12244824607284961224"; if (isDvisibleBy12(num)) cout << "Yes" << endl; else cout << "No" << endl; return 0; } |
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Java
// Java Program to check if // number is divisible by 12 import java.io.*; class GFG { static boolean isDvisibleBy12(String num) { // if number greater then 3 if (num.length() >= 3) { // fiind last digit int d1 = (int)num.charAt(num.length() - 1); // no is odd if (d1 % 2 != 0) return false; // find second last digit int d2 = (int)num.charAt(num.length() - 2); // find sum of all digits int sum = 0; for (int i = 0; i < num.length(); i++) sum += num.charAt(i); return (sum % 3 == 0 && (d2 * 10 + d1) % 4 == 0); } else { // if number is less then // or equal to 100 int number = Integer.parseInt(num); return (number % 12 == 0); } // driver function } public static void main (String[] args) { String num = "12244824607284961224"; if (isDvisibleBy12(num)) System.out.print("Yes"); else System.out.print("No"); } } // This code is contributed by Gitanjali. |
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Python3
# Python Program to check if # number is divisible by 12 import math def isDvisibleBy12( num): # if number greater then 3 if (len(num) >= 3): # fiind last digit d1 = int(num[len(num) - 1]); # no is odd if (d1 % 2 != 0): return False # find second last digit d2 = int(num[len(num) - 2]) # find sum of all digits sum = 0 for i in range(0, len(num) ): sum += int(num[i]) return (sum % 3 == 0 and (d2 * 10 + d1) % 4 == 0) else : # f number is less then # r equal to 100 number = int(num) return (number % 12 == 0) num = "12244824607284961224" if(isDvisibleBy12(num)): print("Yes") else: print("No") # This code is contributed by Gitanjali. |
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C#
// C# Program to check if // number is divisible by 12 using System; class GFG { static bool isDvisibleBy12(string num) { // if number greater then 3 if (num.Length >= 3) { // find last digit int d1 = (int)num[num.Length - 1]; // no is odd if (d1 % 2 != 0) return false; // find second last digit int d2 = (int)num[num.Length - 2]; // find sum of all digits int sum = 0; for (int i = 0; i < num.Length; i++) sum += num[i]; return (sum % 3 == 0 && (d2 * 10 + d1) % 4 == 0); } else { // if number is less then // or equal to 100 int number = int.Parse(num); return (number % 12 == 0); } } // Driver function public static void Main () { String num = "12244824607284961224"; if (isDvisibleBy12(num)) Console.Write("Yes"); else Console.Write("No"); } } // This code is contributed by nitin mittal. |
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PHP
<?php // PHP Program to check if // number is divisible by 12 function isDvisibleBy12($num) { // if number greater then 3 if (strlen($num) >= 3) { // find last digit $d1 = (int)$num[strlen($num) - 1]; // no is odd if ($d1 % 2 != 0) return (0); // find second last digit $d2 = (int)$num[strlen($num) - 2]; // find sum of all digits $sum = 0; for ($i = 0; $i < strlen($num); $i++) $sum += $num[$i]; return ($sum % 3 == 0 && ($d2 * 10 + $d1) % 4 == 0); } else { // if number is less then // or equal to 100 $number = stoi($num); return ($number % 12 == 0); } } // Driver Code $num = "12244824607284961224"; if (isDvisibleBy12($num)) echo("Yes"); else echo("No"); // This code is contributed by Ajit. ?> |
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Output:
Yes
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