Check if a large number is divisible by 11 or not
Given a number, the task is to check if the number is divisible by 11 or not. The input number may be large and it may not be possible to store it even if we use long long int.
Examples:
Input : n = 76945 Output : Yes Input : n = 1234567589333892 Output : Yes Input : n = 363588395960667043875487 Output : No
Since input number may be very large, we cannot use n % 11 to check if a number is divisible by 11 or not, especially in languages like C/C++. The idea is based on following fact.
A number is divisible by 11 if difference of following two is divisible by 11.
- Sum of digits at odd places.
- Sum of digits at even places.
Illustration:
For example, let us consider 76945 Sum of digits at odd places : 7 + 9 + 5 Sum of digits at even places : 6 + 4 Difference of two sums = 21 - 10 = 11 Since difference is divisible by 11, the number 7945 is divisible by 11.
How does this work?
Let us consider 7694, we can write it as 7694 = 7*1000 + 6*100 + 9*10 + 4 The proof is based on below observation: Remainder of 10i divided by 11 is 1 if i is even Remainder of 10i divided by 11 is -1 if i is odd So the powers of 10 only result in values either 1 or -1. Remainder of "7*1000 + 6*100 + 9*10 + 4" divided by 11 can be written as : 7*(-1) + 6*1 + 9*(-1) + 4*1 The above expression is basically difference between sum of even digits and odd digits.
Below is implementation of above fact :
C++
// C++ program to find if a number is divisible by// 11 or not#include<bits/stdc++.h>using namespace std;// Function to find that number divisible by 11 or notint check(string str){ int n = str.length(); // Compute sum of even and odd digit // sums int oddDigSum = 0, evenDigSum = 0; for (int i=0; i<n; i++) { // When i is even, position of digit is odd if (i%2 == 0) oddDigSum += (str[i]-'0'); else evenDigSum += (str[i]-'0'); } // Check its difference is divisible by 11 or not return ((oddDigSum - evenDigSum) % 11 == 0);}// Driver codeint main(){ string str = "76945"; check(str)? cout << "Yes" : cout << "No "; return 0;} |
Java
// Java program to find if a number is// divisible by 11 or notclass IsDivisible{ // Function to find that number divisible by 11 or not static boolean check(String str) { int n = str.length(); // Compute sum of even and odd digit // sums int oddDigSum = 0, evenDigSum = 0; for (int i=0; i<n; i++) { // When i is even, position of digit is odd if (i%2 == 0) oddDigSum += (str.charAt(i)-'0'); else evenDigSum += (str.charAt(i)-'0'); } // Check its difference is divisible by 11 or not return ((oddDigSum - evenDigSum) % 11 == 0); } // main function public static void main (String[] args) { String str = "76945"; if(check(str)) System.out.println("Yes"); else System.out.println("No"); }} |
Python3
# Python 3 code program to find if a number# is divisible by 11 or not# Function to find that number divisible by# 11 or notdef check(st) : n = len(st) # Compute sum of even and odd digit # sums oddDigSum = 0 evenDigSum = 0 for i in range(0,n) : # When i is even, position of digit is odd if (i % 2 == 0) : oddDigSum = oddDigSum + ((int)(st[i])) else: evenDigSum = evenDigSum + ((int)(st[i])) # Check its difference is divisible by 11 or not return ((oddDigSum - evenDigSum) % 11 == 0)# Driver codest = "76945"if(check(st)) : print( "Yes")else : print("No ") # This code is contributed by Nikita tiwari. |
C#
// C# program to find if a number is// divisible by 11 or notusing System;class GFG{ // Function to find that number // divisible by 11 or not static bool check(string str) { int n = str.Length; // Compute sum of even and odd digit // sums int oddDigSum = 0, evenDigSum = 0; for (int i = 0; i < n; i++) { // When i is even, position of // digit is odd if (i % 2 == 0) oddDigSum += (str[i] - '0'); else evenDigSum += (str[i] - '0'); } // Check its difference is // divisible by 11 or not return ((oddDigSum - evenDigSum) % 11 == 0); } // main function public static void Main () { String str = "76945"; if(check(str)) Console.WriteLine("Yes"); else Console.WriteLine("No"); }}// This code is contributed by vt_m. |
PHP
<?php// PHP program to find if a// number is divisible by// 11 or not// Function to find that number// divisible by 11 or notfunction check($str){ $n = strlen($str); // Compute sum of even // and odd digit sums $oddDigSum = 0; $evenDigSum = 0; for ($i = 0; $i < $n; $i++) { // When i is even, position // of digit is odd if ($i % 2 == 0) $oddDigSum += ($str[$i] - '0'); else $evenDigSum += ($str[$i] - '0'); } // Check its difference // is divisible by 11 or not return (($oddDigSum - $evenDigSum) % 11 == 0);}// Driver code$str = "76945";$x = check($str)? "Yes" : "No ";echo($x);// This code is contributed by Ajit.?> |
Javascript
<script>// JavaScript program for the above approach // Function to find that number // divisible by 11 or not function check(str) { let n = str.length; // Compute sum of even and odd digit // sums let oddDigSum = 0, evenDigSum = 0; for (let i = 0; i < n; i++) { // When i is even, position of // digit is odd if (i % 2 == 0) oddDigSum += (str[i] - '0'); else evenDigSum += (str[i] - '0'); } // Check its difference is // divisible by 11 or not return ((oddDigSum - evenDigSum) % 11 == 0); } // Driver Code let str = "76945"; if(check(str)) document.write("Yes"); else document.write("No"); // This code is contributed by chinmoy1997pal.</script> |
Output:
Yes
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