Number is divisible by 29 or not

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

Examples :

```Input : 363927598
Output : No

Input : 292929002929
Output : Yes
```

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 iff 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

 ` `

Output :

```Yes
```