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# 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```
Recommended Practice

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.

1. Sum of digits at odd places.
2. 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 the implementation of above approach:

## C++

 `// C++ program to find if a number is divisible by` `// 11 or not` `#include` `using` `namespace` `std;`   `// Function to find that number divisible by 11 or not` `int` `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

## Java

 `// Java program to find if a number is` `// divisible by 11 or not` `import` `java.io.*;` `class` `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

## 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 not` `def` `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 code` `st ``=` `"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 not` `using` `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

 ``

## Javascript

 ``

Output

`Yes`

Time Complexity: O(n), where n is the given number.
Auxiliary Space: O(1), as we are not using any extra space.

Method: Checking given number is divisible by 11 or not by using the modulo division operator “%”.

## C++

 `// C++ code to check whether` `// the given number is divisible by 11 or not` `#include ` `using` `namespace` `std;`   `int` `main()` `{` `  `  `    ``// input` `    ``long` `long` `n = 1234567589333892;`   `    ``// the above input can also be given as n=input() ->` `    ``// taking input from user finding given number is` `    ``// divisible by 11 or not` `    ``if` `(n % 11 == 0)` `        ``cout << ``"Yes"` `<< endl;` `    ``else` `        ``cout << ``"No"` `<< endl;` `}`   `// This code is contributed by phasing17`

## Java

 `// Java code to check whether` `// the given number is divisible by 11 or not`   `import` `java.io.*;`   `class` `GFG {` `    ``public` `static` `void` `main(String[] args)` `    ``{`   `        ``// input` `        ``Long n = Long.parseUnsignedLong(``"1234567589333892"``);`   `        `  `        ``//finding given number is` `        ``// divisible by 11 or not` `        ``if` `(n % ``11` `== ``0``)` `            ``System.out.println(``"Yes"``);` `        ``else` `            ``System.out.println(``"No"``);` `    ``}` `}`   `// This code is contributed by phasing17`

## Python3

 `# Python code ` `# To check whether the given number is divisible by 11 or not`   `#input ` `n``=``1234567589333892` `# the above input can also be given as n=input() -> taking input from user` `# finding given number is divisible by 11 or not` `if` `int``(n)``%``11``=``=``0``:` `  ``print``(``"Yes"``) ` `else``: ` `  ``print``(``"No"``)`   `  ``# this code is contributed by gangarajula laxmi`

## C#

 `// C# code to check whether` `// the given number is divisible by 11 or not` `using` `System;`   `class` `GFG {` `    ``public` `static` `void` `Main(``string``[] args)` `    ``{`   `        ``// input` `        ``long` `n = 1234567589333892;`   `        ``// the above input can also be given as n=input() ->` `        ``// taking input from user finding given number is` `        ``// divisible by 11 or not` `        ``if` `(n % 11 == 0)` `            ``Console.WriteLine(``"Yes"``);` `        ``else` `            ``Console.WriteLine(``"No"``);` `    ``}` `}`   `// This code is contributed by phasing17`

## Javascript

 `// JavaScript code to check whether ` `// the given number is divisible by 11 or not`   `// input ` `let n = 1234567589333892`   `// the above input can also be given as n=input() -> taking input from user` `// finding given number is divisible by 11 or not` `if` `(n % 11 == 0)` `  ``console.log(``"Yes"``) ` `else` `  ``console.log(``"No"``)`   `// this code is contributed by phasing17`

## PHP

 ``

Output

`Yes`

Time Complexity: O(1) because it is performing constant operations
Auxiliary Space: O(1)

Method: Checking given number is divisible by 11 or not using modulo division.

## C++

 `// C++ program to check if given number is divisible by 11` `// or not using modulo division`   `#include ` `using` `namespace` `std;`   `int` `main()` `{`   `    ``// input number` `    ``int` `num = 76945;` `    ``// checking if the given number is divisible by 11 or` `    ``// not using modulo division operator if the output of` `    ``// num%11 is equal to 0 then given number is divisible` `    ``// by 11 otherwise not divisible by 11` `    ``if` `(num % 11 == 0) {` `        ``cout << ``" divisible"``;` `    ``}` `    ``else` `{` `        ``cout << ``" not divisible"``;` `    ``}` `    ``return` `0;` `}`   `// this code is contributed by gangarajula laxmi`

## Java

 `// java program to check if given number is divisible by 11` `// or not using modulo division`   `import` `java.io.*;`   `class` `GFG {` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``// input number` `        ``int` `num = ``76945``;` `        ``// checking if the given number is divisible by 11` `        ``// or not` `        ``// using modulo division operator if the output of` `        ``// num%11 is equal to 0 then given number is` `        ``// divisible by 11 otherwise not divisible by 11` `        ``if` `(num % ``11` `== ``0``) {` `            ``System.out.println(``" divisible"``);` `        ``}` `        ``else` `{` `            ``System.out.println(``" not divisible"``);` `        ``}` `    ``}` `}`   `// this code is contributed by gangarajula laxmi`

## Python3

 `# Python3 code for the above approach`   `# To check whether the given number is divisible by 11 or not`   `# input ` `n ``=` `76945` `        `  `# finding given number is divisible by 11 or not` `if` `(n ``%` `11` `=``=` `0``):` `    ``print``(``"Yes"``)` `else``:` `    ``print``(``"No"``)`   `#  This code is contributed by phasing17`

## C#

 `using` `System;`   `public` `class` `GFG {`   `    ``static` `public` `void` `Main()` `    ``{`   `        ``// input number` `        ``double` `num = 76945;` `      `  `        ``// checking if the given number is divisible by 11` `        ``// or not using modulo division operator if the` `        ``// output of num%11 is equal to 0 then given number` `        ``// is divisible by 11 otherwise not divisible by 11` `        ``if` `(num % 11 == 0) {` `            ``Console.Write(``" divisible"``);` `        ``}` `        ``else` `{` `            ``Console.Write(``" not divisible"``);` `        ``}` `    ``}` `  `  `  ``// this code is contributed by gangarajula laxmi`

## Javascript

 ``

## PHP

 ``

Output

` divisible`

Time Complexity : O(1)

Space Complexity : O(1)

Method 4:

1. Initialize two variables: alternating_sum to store the alternating sum of the digits and multiplier to keep track of whether to add or subtract each digit. Set alternating_sum to 0 and multiplier to 1.

2. Use a while loop to iterate over the digits of the number num. The loop will continue as long as num is greater than 0.

a. Get the last digit of the number by using the modulo operator (%) with 10. Store this digit in a temporary variable.

b. Update the alternating_sum by adding (or subtracting) the last digit, multiplied by multiplier.

c. Change the value of multiplier to its opposite by multiplying it by -1.

d. Remove the last digit of the number by using integer division (//) with 10.

3. After the loop, check if the alternating_sum is divisible by 11 by using the modulo operator (%) with 11. If it is, return True, otherwise return False.

## C++

 `#include `   `bool` `is_divisible_by_11(``int` `num) {` `    ``int` `alternating_sum = 0;` `    ``int` `multiplier = 1;` `    ``while` `(num > 0) {` `        ``alternating_sum += multiplier * (num % 10);` `        ``multiplier *= -1;` `        ``num /= 10;` `    ``}` `    ``// checking if divisible by 11 or not` `    ``return` `alternating_sum % 11 == 0;` `}`   `int` `main() {` `    ``std::cout << std::boolalpha; ``// enable printing of true/false instead of 1/0` `    ``std::cout << is_divisible_by_11(11) << std::endl; ``// true` `    ``std::cout << is_divisible_by_11(22) << std::endl; ``// true` `    ``std::cout << is_divisible_by_11(121) << std::endl; ``// true` `    ``std::cout << is_divisible_by_11(10) << std::endl; ``// false` `    ``return` `0;` `}`

## Python3

 `def` `is_divisible_by_11(num):` `    ``alternating_sum ``=` `0` `    ``multiplier ``=` `1` `    ``while` `num > ``0``:` `        ``alternating_sum ``+``=` `multiplier ``*` `(num ``%` `10``)` `        ``multiplier ``*``=` `-``1` `        ``num ``/``/``=` `10` `    ``#checking if divisible by 11 or not` `    ``return` `alternating_sum ``%` `11` `=``=` `0`   `print``(is_divisible_by_11(``11``)) ``# True` `print``(is_divisible_by_11(``22``)) ``# True` `print``(is_divisible_by_11(``121``)) ``# True` `print``(is_divisible_by_11(``10``)) ``# False`

## Java

 `public` `class` `GFG {` `    ``public` `static` `boolean` `isDivisibleBy11(``int` `num)` `    ``{` `        ``int` `alternatingSum = ``0``;` `        ``int` `multiplier = ``1``;` `        ``while` `(num > ``0``) {` `            ``alternatingSum += multiplier * (num % ``10``);` `            ``multiplier *= -``1``;` `            ``num /= ``10``;` `        ``}` `        ``// Checking if divisible by 11 or not` `        ``return` `alternatingSum % ``11` `== ``0``;` `    ``}`   `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``System.out.println(isDivisibleBy11(``11``)); ``// true` `        ``System.out.println(isDivisibleBy11(``22``)); ``// true` `        ``System.out.println(isDivisibleBy11(``121``)); ``// true` `        ``System.out.println(isDivisibleBy11(``10``)); ``// false` `    ``}` `}`

## C#

 `using` `System;`   `class` `Program {` `    ``static` `bool` `IsDivisibleBy11(``int` `num)` `    ``{` `        ``int` `alternatingSum = 0;` `        ``int` `multiplier = 1;` `        ``while` `(num > 0) {` `            ``alternatingSum += multiplier * (num % 10);` `            ``multiplier *= -1;` `            ``num /= 10;` `        ``}` `        ``// checking if divisible by 11 or not` `        ``return` `alternatingSum % 11 == 0;` `    ``}`   `    ``static` `void` `Main()` `    ``{` `        ``Console.WriteLine(IsDivisibleBy11(11)); ``// true` `        ``Console.WriteLine(IsDivisibleBy11(22)); ``// true` `        ``Console.WriteLine(IsDivisibleBy11(121)); ``// true` `        ``Console.WriteLine(IsDivisibleBy11(10)); ``// false` `    ``}` `}`

## Javascript

 `function` `is_divisible_by_11(num) ` `{`   `    ``// Initialize an integer named "alternating_sum" to 0` `    ``let alternating_sum = 0;` `    `  `    ``// Initialize an integer named "multiplier" to 1` `    ``let multiplier = 1;` `    `  `    ``// Loop through the digits of the input integer from right to left` `    ``while` `(num > 0) ` `    ``{` `    `  `        ``// Add the product of the current digit and "multiplier" to "alternating_sum"` `        ``alternating_sum += multiplier * (num % 10);` `        `  `        ``// Toggle the value of "multiplier" between 1 and -1` `        ``multiplier *= -1;` `        `  `        ``// Remove the last digit from the input integer` `        ``num = Math.floor(num / 10);` `    ``}` `    ``// Check if "alternating_sum" is divisible by 11` `    ``return` `alternating_sum % 11 == 0;` `}`   `// Enable printing of true/false instead of 1/0` `console.log(is_divisible_by_11(11)); ``// true` `console.log(is_divisible_by_11(22)); ``// true` `console.log(is_divisible_by_11(121)); ``// true` `console.log(is_divisible_by_11(10)); ``// false`

Output

```True
True
True
False```

Time complexity: O(log(n))
Auxiliary space:  O(1)

### Method: Using string manipulation

• The input number is converted to a string, and then the digits are processed one by one in reverse order using a loop.
• The alternating sum is computed by adding each digit to the sum with the appropriate sign (+ or -) based on its position.
• Finally, the alternating sum is checked for divisibility by 11 using the modulo operator (%).

## Python3

 `def` `is_divisible_by_11(number):` `    ``# Compute the alternating sum of the digits from right to left` `    ``alternating_sum ``=` `0` `    ``for` `i, digit ``in` `enumerate``(``reversed``(``str``(number))):` `        ``alternating_sum ``+``=` `(``-``1``) ``*``*` `i ``*` `int``(digit)`   `    ``# If the alternating sum is divisible by 11, then the number is divisible by 11` `    ``return` `alternating_sum ``%` `11` `=``=` `0`   `st ``=` `"76945"` `if``(is_divisible_by_11(st)) :` `    ``print``( ``"Yes"``)` `else` `:` `    ``print``(``"No "``) ` `   `

Output

`Yes`

Time complexity:

The time complexity of the function is O(n), where n is the number of digits in the input number.
This is because the function processes each digit of the input number once in the loop.
Auxiliary space:

The auxiliary space complexity of the function is O(1), because it uses a constant amount of extra memory to store the alternating sum and loop variables.
The amount of memory used does not depend on the size of the input number.

This article is contributed by DANISH_RAZA . If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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