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# Number is divisible by 29 or not

• Difficulty Level : Easy
• Last Updated : 22 Feb, 2023

Given a large number n, find if the number is divisible by 29.
Examples :

```Input : 363927598
Output : No

Input : 292929002929
Output : Yes```

A quick solution to check if a number is divisible by 29 or not is to add 3 times of last digit to rest number and repeat this process until number comes 2 digit. The given number is divisible by 29 if the obtained two digit number is divisible by 29.
Number is 348,
Three times of last digit + Rest of the number = 8*3 + 34 = 58
Since 58 is divisible by 29, 348 is also divisible by 29.

## C++

 `// CPP program to demonstrate above method` `// to check divisibility by 29.` `#include ` `using` `namespace` `std;`   `// Returns true if n is divisible by 29` `// else returns false.` `bool` `isDivisible(``long` `long` `int` `n)` `{` `    ``// add the lastdigit*3 to renaming ` `    ``// number until number comes only` `    ``// 2 digit` `    ``while` `(n / 100) ` `    ``{` `        ``int` `last_digit = n % 10;` `        ``n /= 10;` `        ``n += last_digit * 3;` `    ``}`   `    ``// return true if number is` `    ``// divisible by 29 another` `    ``return` `(n % 29 == 0);` `}`   `// Driver Code` `int` `main()` `{` `    ``long` `long` `int` `n = 348;` `    ``if` `(isDivisible(n))` `        ``cout << ``"Yes"` `<< endl;` `    ``else` `        ``cout << ``"No"` `<< endl;` `    ``return` `0;` `}`

## Java

 `// Java program to demonstrate above method` `// to check divisibility by 29.`   `import` `java.io.*;`   `class` `GFG {` `    `  `    ``// Returns true if n is divisible by 29` `    ``// else returns false.` `    ``static` `boolean` `isDivisible(``long` `n)` `    ``{` `        `  `        ``// add the lastdigit*3 to renaming` `        ``// number until number comes only` `        ``// 2 digit` `        ``while` `(n / ``100` `> ``0``) {` `            `  `            ``int` `last_digit = (``int``)n % ``10``;` `            ``n /= ``10``;` `            ``n += last_digit * ``3``;` `        ``}`   `        ``// return true if number is` `        ``// divisible by 29 another` `        ``return` `(n % ``29` `== ``0``);` `    ``}` `    `  `    ``// Driver code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``long` `n = ``348``;` `        `  `        ``if` `(isDivisible(n))` `            ``System.out.println(``"Yes"``);` `        ``else` `            ``System.out.println(``"No"``);` `    ``}` `}`   `// This code is contributed by vt_m.`

## Python3

 `# Python3 program to demonstrate above ` `# method to check divisibility by 29.`   `# Returns true if n is divisible ` `# by 29 else returns false.` `def` `isDivisible(n):`   `    ``# add the lastdigit*3 to renaming ` `    ``# number until number comes only` `    ``# 2 digit` `    ``while` `(``int``(n ``/` `100``)) :` `        ``last_digit ``=` `int``(n ``%` `10``)` `        ``n ``=` `int``(n ``/` `10``)` `        ``n ``+``=` `last_digit ``*` `3` `    `  `    ``# return true if number is` `    ``# divisible by 29 another` `    ``return` `(n ``%` `29` `=``=` `0``)`   `# Driver Code` `n ``=` `348`   `if``(isDivisible(n) !``=` `0``):` `    ``print``(``"Yes"``)` `else``:` `    ``print``(``"No"``)`   `# This code is contributed by Smitha Dinesh Semwal.`

## C#

 `// C# program to demonstrate above method` `// to check divisibility by 29.` `using` `System;` `class` `GFG ` `{` `    `  `    ``// Returns true if n is divisible by 29` `    ``// else returns false.` `    ``static` `bool` `isDivisible(``long` `n)` `    ``{` `        `  `        ``// add the lastdigit*3 to renaming` `        ``// number until number comes only` `        ``// 2 digit` `        ``while` `(n / 100 > 0) ` `        ``{` `            `  `            ``int` `last_digit = (``int``)n % 10;` `            ``n /= 10;` `            ``n += last_digit * 3;` `        ``}`   `        ``// return true if number is` `        ``// divisible by 29 another` `        ``return` `(n % 29 == 0);` `    ``}` `    `  `    ``// Driver code` `    ``public` `static` `void` `Main()` `    ``{` `        ``long` `n = 348;` `        `  `        ``if` `(isDivisible(n))` `            ``Console.Write(``"Yes"``);` `        ``else` `            ``Console.Write(``"No"``);` `    ``}` `}`   `// This code is contributed by nitin mittal`

## PHP

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## Javascript

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Output :

`Yes`

Time Complexity: O(n) where n is given number.

Space Complexity: O(1) as we are not using any extra space.

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