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# To check divisibility of any large number by 999

You are given an n-digit large number, you have to check whether it is divisible by 999 without dividing or finding modulo of number by 999.

Examples:

```Input : 235764
Output : Yes

Input : 23576
Output : No```

Since input number may be very large, we cannot use n % 999 to check if a number is divisible by 999 or not, especially in languages like C/C++. The idea is based on following fact.

Recommended Practice

The solutions is based on below fact.

A number is divisible by 999 if sum of its 3-digit-groups (if required groups are formed by appending a 0s at the beginning) is divisible by 999.

Illustration:

```Input : 235764
Output : Yes
Explanation : Step I - read input : 235, 764
Step II - 235 + 764 = 999
As result is 999 then we can
conclude that it is divisible by 999.

Input : 1244633121
Output : Yes
Explanation : Step I - read input : 1, 244, 633, 121
Step II - 001 + 244 + 633 + 121 = 999
As result is 999 then we can conclude
that it is divisible by 999.

Input : 999999999
Output : Yes
Explanation : Step I - read input : 999, 999, 999
Step II - 999 + 999 + 999 = 2997
Step III - 997 + 002 = 999
As result is 999 then we can conclude
that it is divisible by 999.```

How does this work?

```Let us consider 235764, we can write it as
235764 = 2*105 + 3*104 + 5*103 +
7*102 + 6*10 + 4

The idea is based on below observation:
Remainder of 103 divided by 999 is 1
For i > 3, 10i % 999 = 10i-3 % 999

Let us see how we use above fact.
Remainder of 2*105 + 3*104 + 5*103 +
7*102 + 6*10 + 4
Remainder with 999 can be written as :
2*100 + 3*10 + 5*1 + 7*100 + 6*10 + 4
The above expression is basically sum of
groups of size 3.

Since the sum is divisible by 999, answer is yes.```

A simple and efficient method is to take input in form of string (make its length in form of 3*m by adding 0 to left of number if required) and then you have to add the digits in blocks of three from right to left until it become a 3 digit number and if that result is 999 we can say that number is divisible by 999.

As in the case of “divisibility by 9” we check that sum of all digit is divisible by 9 or not, the same thing follows within the case of divisibility by 999. We sum up all 3-digits group from right to left and check whether the final result is 999 or not.

Implementation:

## Java

 `//Java for divisibility of number by 999`   `class` `Test` `{    ` `    ``// Method to check divisibility` `    ``static` `boolean` `isDivisible999(String num)` `    ``{` `        ``int` `n = num.length();` `        ``if` `(n == ``0` `&& num.charAt(``0``) == ``'0'``)` `           ``return` `true``;` `     `  `        ``// Append required 0s at the beginning.` `        ``if` `(n % ``3` `== ``1``)` `           ``num = ``"00"` `+ num;` `        ``if` `(n % ``3` `== ``2``)` `           ``num = ``"0"` `+ num;` `     `  `        ``// add digits in group of three in gSum` `        ``int` `gSum = ``0``;` `        ``for` `(``int` `i = ``0``; i ``1000``)` `        ``{` `            ``num = Integer.toString(gSum);` `            ``n = num.length();` `            ``gSum = isDivisible999(num) ? ``1` `: ``0``;` `        ``}` `        ``return` `(gSum == ``999``);` `    ``}` `    `  `    ``// Driver method` `    ``public` `static` `void` `main(String args[]) ` `    ``{` `        ``String num = ``"1998"``;` `     `  `        ``System.out.println(isDivisible999(num) ? ``"Divisible"` `: ``"Not divisible"``);` `    ``}` `}`

## Python 3

 `# Python3 program for divisibility ` `# of number by 999 `   `# function to check divisibility ` `def` `isDivisible999(num):` `    ``n ``=` `len``(num);` `    ``if``(n ``=``=` `0` `or` `num[``0``] ``=``=` `'0'``):` `        ``return` `true`   `    ``# Append required 0s at the beginning.` `    ``if``((n ``%` `3``) ``=``=` `1``):` `        ``num ``=` `"00"` `+` `num` `    ``if``((n ``%` `3``) ``=``=` `2``):` `        ``num ``=` `"0"` `+` `num`   `    ``# add digits in group of three in gSum     ` `    ``gSum ``=` `0` `    ``for` `i ``in` `range``(``0``, n, ``3``):` `        `  `        ``# group saves 3-digit group ` `        ``group ``=` `0` `        ``group ``+``=` `(``ord``(num[i]) ``-` `48``) ``*` `100` `        ``group ``+``=` `(``ord``(num[i ``+` `1``]) ``-` `48``) ``*` `10` `        ``group ``+``=` `(``ord``(num[i ``+` `2``]) ``-` `48``)` `        ``gSum ``+``=` `group`   `    ``# calculate result till 3 digit sum     ` `    ``if``(gSum > ``1000``):` `        ``num ``=` `str``(gSum)` `        ``n ``=` `len``(num)` `        ``gSum ``=` `isDivisible999(num)` `    ``return` `(gSum ``=``=` `999``) `   `# Driver code ` `if` `__name__``=``=``"__main__"``:` `    ``num ``=` `"1998"` `    ``n ``=` `len``(num)` `    ``if``(isDivisible999(num)):` `        ``print``(``"Divisible"``)` `    ``else``:` `        ``print``(``"Not divisible"``)` `        `  `# This code is contributed` `# by Sairahul Jella`

## C#

 `// C# code for divisibility of number by 999 `   `using` `System;` `class` `Test ` `{ ` `    ``// Method to check divisibility ` `    ``static` `bool` `isDivisible999(String num) ` `    ``{ ` `        ``int` `n = num.Length; ` `        ``if` `(n == 0 && num[0] == ``'0'``) ` `        ``return` `true``; ` `    `  `        ``// Append required 0s at the beginning. ` `        ``if` `(n % 3 == 1) ` `        ``num = ``"00"` `+ num; ` `        ``if` `(n % 3 == 2) ` `        ``num = ``"0"` `+ num; ` `    `  `        ``// add digits in group of three in gSum ` `        ``int` `gSum = 0; ` `        ``for` `(``int` `i = 0; i 1000) ` `        ``{ ` `            ``num = Convert.ToString(gSum); ` `            ``n = num.Length ; ` `            ``gSum = isDivisible999(num) ? 1 : 0; ` `        ``} ` `        ``return` `(gSum == 999); ` `    ``} ` `    `  `    ``// Driver method ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``String num = ``"1998"``; ` `    `  `        ``Console.WriteLine(isDivisible999(num) ? ``"Divisible"` `: ``"Not divisible"``); ` `    ``} ` `    ``// This code is contributed by Ryuga` `} `

## Javascript

 ``

Output

`Divisible`

Time complexity : O(n)
Auxiliary Space : O(1)

Method 2: Checking given any number is divisible by 999 or not by using modulo division operator “%”.

Implementation:

## C++

 `#include ` `using` `namespace` `std;` `int` `main()` `{` `    ``//input` `    ``long` `long` `int` `n=235764;` `     `  `     `  `    ``// finding given number is divisible by 999 or not` `    ``if` `(n%999==0)` `    ``{` `        ``cout << ``"Yes"``;` `    ``}` `    ``else` `    ``{` `        ``cout << ``"No"``;` `    ``}` `   `  `    ``return` `0;` `}`   `// This code is contributed by satwik4409.`

## Java

 `/*package whatever //do not write package name here */` `import` `java.io.*;`   `class` `GFG {` `  ``public` `static` `void` `main (String[] args) {`   `    ``// input ` `    ``long` `n = ``235764``;`   `    ``// finding given number is divisible by 999 or not` `    ``if` `(n % ``999` `== ``0``)` `      ``System.out.println(``"Yes"``);` `    ``else` `      ``System.out.println(``"No"``); `   `  ``}` `}`   `// This code is contributed by ksrikanth0498.`

## Python3

 `# Python code ` `# To check whether the given number is divisible by 999 or not`   `#input ` `n``=``235764` `# the above input can also be given as n=input() -> taking input from user` `# finding given number is divisible by 999 or not` `if` `int``(n)``%``999``=``=``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 999 or not` `using` `System;`   `public` `class` `GFG{`   `    ``static` `public` `void` `Main ()` `    ``{` `      `  `       ``// input ` `       ``long` `n = 235764;` `      `  `        ``// finding given number is divisible by 999 or not` `       ``if` `(n % 999 == 0)` `            ``Console.Write(``"Yes"``);` `        ``else` `            ``Console.Write(``"No"``); ` `    ``} ` `}`   `// This code is contributed by ksrikanth0498`

## PHP

 ``

## Javascript

 ``

Output

`Yes`

Time Complexity : O(1)

Auxiliary Space: O(1)

Method 3: The function returns a Boolean value indicating whether the input integer is divisible by 999 or not. The function calculates this by checking if the sum of the digits of num is divisible by 9 and if the last three digits of num are divisible by 27. This code then tests the function with the examples: 999, 998 and 9987 and outputs the values True, False and False respectively.

## C++

 `convert C++ code into Javascript code`   `#include ` `using` `namespace` `std;`   `bool` `is_divisible_by_999(``int` `num) {` `    ``string num_str = to_string(num);` `    ``int` `sum = 0;` `    ``for` `(``char` `digit : num_str) {` `        ``sum += digit - ``'0'``;` `    ``}` `    ``return` `sum % 9 == 0 && stoi(num_str.substr(num_str.length() - 3)) % 27 == 0;` `}`   `int` `main() {` `    ``cout << is_divisible_by_999(999) << endl;  ``// True` `    ``cout << is_divisible_by_999(998) << endl;  ``// False` `    ``cout << is_divisible_by_999(9987) << endl;  ``// False` `    ``return` `0;` `}`

## Java

 `public` `class` `Main {` `    ``public` `static` `boolean` `isDivisibleBy999(``int` `num) {` `        ``String numStr = String.valueOf(num);` `        ``int` `sum = ``0``;` `        ``for` `(``int` `i = ``0``; i < numStr.length(); i++) {` `            ``char` `digit = numStr.charAt(i);` `            ``sum += Character.getNumericValue(digit);` `        ``}` `        ``return` `sum % ``9` `== ``0` `&& Integer.parseInt(numStr.substring(numStr.length() - ``3``)) % ``27` `== ``0``;` `    ``}`   `    ``public` `static` `void` `main(String[] args) {` `        ``System.out.println(isDivisibleBy999(``999``));  ``// true` `        ``System.out.println(isDivisibleBy999(``998``));  ``// false` `        ``System.out.println(isDivisibleBy999(``9987``)); ``// false` `    ``}` `}`

## Python3

 `def` `is_divisible_by_999(num):` `    ``num_str ``=` `str``(num)` `    ``return` `sum``(``int``(digit) ``for` `digit ``in` `num_str) ``%` `9` `=``=` `0` `and` `int``(num_str[``-``3``:]) ``%` `27` `=``=` `0`   `print``(is_divisible_by_999(``999``)) ``# True` `print``(is_divisible_by_999(``998``)) ``# False` `print``(is_divisible_by_999(``9987``)) ``# False`

## C#

 `using` `System;`   `class` `Program {` `    ``static` `bool` `IsDivisibleBy999(``int` `num) {` `        ``string` `numStr = num.ToString();` `        ``int` `digitSum = 0;` `        ``foreach` `(``char` `digitChar ``in` `numStr) {` `            ``digitSum += digitChar - ``'0'``;` `        ``}` `        ``return` `digitSum % 9 == 0 && ``int``.Parse(numStr.Substring(numStr.Length - 3)) % 27 == 0;` `    ``}`   `    ``static` `void` `Main() {` `        ``Console.WriteLine(IsDivisibleBy999(999)); ``// True` `        ``Console.WriteLine(IsDivisibleBy999(998)); ``// False` `        ``Console.WriteLine(IsDivisibleBy999(9987)); ``// False` `    ``}` `}`

## Javascript

 `function` `is_divisible_by_999(num) {` `    ``// Convert the input number to a string` `    ``var` `num_str = num.toString();`   `    ``// Initialize a variable to hold the sum of the digits` `    ``var` `sum = 0;`   `    ``// Iterate over each character in the string representation of the number` `    ``for` `(``var` `i = 0; i < num_str.length; i++) {` `        ``// Add the numeric value of the current digit to the sum variable` `        ``sum += parseInt(num_str[i]);` `    ``}`   `    ``// Check if the sum is divisible by 9 and the last three digits of the number are divisible by 27` `    ``return` `(sum % 9 == 0 && parseInt(num_str.substring(num_str.length - 3)) % 27 == 0);` `}`   `// Test the function with some sample inputs` `console.log(is_divisible_by_999(999));   ``// true` `console.log(is_divisible_by_999(998));   ``// false` `console.log(is_divisible_by_999(9987));  ``// false`   `// This code is contributed by bhardwajji.`

Output

```True
False
False```

Time complexity: O(1),
Auxiliary Space : O(1)

### Method 4: Using string manipulation

1. Convert the number to a string.
2. Find the sum of every third digit from the right end to the left end.
3. If the sum is divisible by 3, then the original number is divisible by 999.

## Java

 `public` `class` `GFG {` `    ``public` `static` `boolean` `isDivisibleBy999(``int` `n) {` `        ``String s = Integer.toString(n);` `        ``int` `l = s.length();` `        ``int` `sum = ``0``;` `        ``for` `(``int` `i = l - ``1``; i >= ``0``; i -= ``3``) {` `            ``int` `j = Math.max(i - ``2``, -``1``);` `            ``sum += Integer.parseInt(s.substring(j + ``1``, i + ``1``));` `        ``}` `        ``return` `sum % ``3` `== ``0``;` `    ``}`   `    ``public` `static` `void` `main(String[] args) {` `        ``int` `n1 = ``998``;` `        ``int` `n2 = ``999``;` `        ``System.out.println(isDivisibleBy999(n1));` `        ``System.out.println(isDivisibleBy999(n2));` `    ``}` `}`

## Python3

 `# python code to check if given number is divisible by 998 or not` `def` `is_divisible_by_999(n):` `    ``s ``=` `str``(n)` `    ``l ``=` `len``(s)` `    ``sum` `=` `0` `    ``for` `i ``in` `range``(l``-``1``, ``-``1``, ``-``3``):` `        ``j ``=` `max``(i``-``2``, ``-``1``)` `        ``sum` `+``=` `int``(s[j``+``1``:i``+``1``])` `    ``return` `sum` `%` `3` `=``=` `0`   `# Example usage` `n1 ``=` `998` `n2 ``=` `999` `print``(is_divisible_by_999(n1))  ``# Output: True` `print``(is_divisible_by_999(n2))  ``# Output: False`

## C++

 `#include ` `#include `   `using` `namespace` `std;`   `bool` `isDivisibleBy999(``int` `n)` `{` `    ``string s = to_string(n);` `    ``int` `l = s.length();` `    ``int` `sum = 0;` `    ``for` `(``int` `i = l - 1; i >= 0; i -= 3) {` `        ``int` `j = max(i - 2, -1);` `        ``sum += stoi(s.substr(j + 1, i - j));` `    ``}` `    ``return` `sum % 3 == 0;` `}`   `int` `main()` `{` `    ``int` `n1 = 998;` `    ``int` `n2 = 999;` `    ``cout << isDivisibleBy999(n1) << endl;` `    ``cout << isDivisibleBy999(n2) << endl;` `    ``return` `0;` `}`

## C#

 `using` `System;`   `public` `class` `GFG {` `    ``public` `static` `bool` `IsDivisibleBy999(``int` `n) {` `        ``string` `s = n.ToString();` `        ``int` `l = s.Length;` `        ``int` `sum = 0;` `        ``for` `(``int` `i = l - 1; i >= 0; i -= 3) {` `            ``int` `j = Math.Max(i - 2, -1);` `            ``sum += ``int``.Parse(s.Substring(j + 1, i - j));` `        ``}` `        ``return` `sum % 3 == 0;` `    ``}`   `    ``public` `static` `void` `Main(``string``[] args) {` `        ``int` `n1 = 998;` `        ``int` `n2 = 999;` `        ``Console.WriteLine(IsDivisibleBy999(n1));` `        ``Console.WriteLine(IsDivisibleBy999(n2));` `    ``}` `}` `// This code is contributed by shiv1o43g`

## Javascript

 `// JavaScript code to check if given number is divisible by 998 or not` `function` `isDivisibleBy998(n) {` `  ``let s = n.toString();` `  ``let l = s.length;` `  ``let sum = 0;` `  ``for` `(let i = l - 1; i >= 0; i -= 3) {` `    ``let j = Math.max(i - 2, -1);` `    ``sum += parseInt(s.substring(j + 1, i + 1));` `  ``}` `  ``return` `sum % 3 === 0;` `}`   `// Example usage` `let n1 = 998;` `let n2 = 999;` `console.log(isDivisibleBy998(n1));  ` `console.log(isDivisibleBy998(n2));  `

Output

```False
True```

The time complexity of the given function is_divisible_by_999 is O(N/3) where N is the number of digits in the input number n. This is because the loop iterates over every third digit of the input number, and performs constant time operations (such as string slicing and integer conversion) on each iteration.

The auxiliary space used by the function is O(1), since it uses a constant amount of extra space to store the loop variables and the running sum.

More Divisibility Algorithms.
This article is contributed by Shivam Pradhan (anuj_charm). 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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