Number is divisible by 29 or not

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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 <iostream>
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

<?php
// PHP program to demonstrate 
// above method to check 
// divisibility by 29.
 
// Returns true if n is 
// divisible by 29
// else returns false.
function isDivisible($n)
{
    // add the lastdigit*3 to 
    // remaining number until 
    // number becomes of only 
    // 2 digit
    while (intval($n / 100)) 
    {
        $last_digit = $n % 10;
        $n = intval($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))
    echo "Yes";
else
    echo "No" ;
 
// This code is contributed by Sam007
?>

Javascript

<script>
// Javascript program to demonstrate 
// above method to check 
// divisibility by 29.
 
// Returns true if n is 
// divisible by 29
// else returns false.
function isDivisible(n)
{
 
    // add the lastdigit*3 to 
    // remaining number until 
    // number becomes of only 
    // 2 digit
    while (parseInt(n / 100)) 
    {
        let last_digit = n % 10;
        n = parseInt(n / 10);
        n += last_digit * 3;
    }
 
    // return true if number is
    // divisible by 29 another
    return (n % 29 == 0);
}
 
// Driver Code
let n = 348;
if (isDivisible(n))
    document.write("Yes");
else
    document.write("No") ;
 
// This code is contributed by _saurabh_jaiswal
</script>

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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