Check if any large number is divisible by 17 or not

Given a number, the task is to quickly check if the number is divisible by 17 or not. 
Example:

Input : x = 34
Output : Yes

Input : x = 47
Output : No

A solution to the problem is to extract the last digit and subtract 5 times of the last digit from the remaining number and repeat this process until a two-digit number is obtained. If the obtained two-digit number is divisible by 17, then the given number is divisible by 17.
Approach:

• Extract the last digit of the number/truncated number every time
• Subtract 5*(last digit of the previous number) from the truncated number
• Repeat the above three steps as long as necessary.

Illustration:

3978-->397-5*8=357-->35-5*7=0. 
So 3978 is divisible by 17.

Mathematical Proof : 
Let be any number such that =100a+10b+c . 
Now assume that is divisible by 17. Then 
0 (mod 17) 
100a+10b+c0 (mod 17) 
10(10a+b)+c0 (mod 17) 
10+c0 (mod 17)
Now that we have separated the last digit from the number, we have to find a way to use it. 
Make the coefficient of 1. 
In other words, we have to find an integer such that n such that 10n1 mod 17. 
It can be observed that the smallest n which satisfies this property is -5 as -501 mod 17. 
Now we can multiply the original equation 10+c0 (mod 17) 
by -5 and simplify it: 
-50-5c0 (mod 17) 
-5c0 (mod 17) 
We have found out that if 0 (mod 17) then, 
-5c0 (mod 17). 
In other words, to check if a 3-digit number is divisible by 17, 
we can just remove the last digit, multiply it by 5, 
and then subtract it from the rest of the two digits. 

Program :

C++

// CPP Program to validate the above logic
#include <bits/stdc++.h>

using namespace std;

// Function to check if the
// number is divisible by 17 or not
bool isDivisible(long long int n)
{

    while (n / 100)
    {
        // Extracting the last digit
        int d = n % 10;

        // Truncating the number
        n /= 10;

        // Subtracting the five times the
        // last digit from the remaining number
        n -= d * 5;
    }

    // Return n is divisible by 17
    return (n % 17 == 0);
}

// Driver code
int main()
{
    long long int n = 19877658;
    if (isDivisible(n))
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
    return 0;
}

Java

// Java Program to validate the above logic

import java.io.*;

class GFG {


// Function to check if the
// number is divisible by 17 or not
 static boolean isDivisible(long n)
{

    while (n / 100>0)
    {
        // Extracting the last digit
        long d = n % 10;

        // Truncating the number
        n /= 10;

        // Subtracting the five times the
        // last digit from the remaining number
        n -= d * 5;
    }

    // Return n is divisible by 17
    return (n % 17 == 0);
}

// Driver code

    public static void main (String[] args) {
    long n = 19877658;
    if (isDivisible(n))
        System.out.println( "Yes");
    else
        System.out.println( "No");
    }
}
// This code is contributed by inder_verma.

Python 3

# Python 3 Program to validate
# the above logic

# Function to check if the
# number is divisible by 17 or not
def isDivisible(n) :

    while(n // 100) :

        # Extracting the last digit
        d = n % 10

        # Truncating the number
        n //= 10

        # Subtracting the five times
        # the last digit from the
        # remaining number
        n -= d * 5

    # Return n is divisible by 17
    return (n % 17 == 0)

# Driver Code
if __name__ == "__main__" :

    n = 19877658

    if isDivisible(n) :
        print("Yes")
    else :
        print("No")

# This code is contributed
# by ANKITRAI1

C#

// C# Program to validate the above logic

using System;

class GFG {


// Function to check if the
// number is divisible by 17 or not
 static bool isDivisible(long n)
{

    while (n / 100>0)
    {
        // Extracting the last digit
        long d = n % 10;

        // Truncating the number
        n /= 10;

        // Subtracting the five times the
        // last digit from the remaining number
        n -= d * 5;
    }

    // Return n is divisible by 17
    return (n % 17 == 0);
}

// Driver code

    public static void Main () {
    long n = 19877658;
    if (isDivisible(n))
        Console.Write( "Yes");
    else
        Console.Write( "No");
    }
}

PHP

<?php
// PHP Program to validate the above logic

// Function to check if the
// number is divisible by 17 or not
function isDivisible($n)
{

    while ($n / 100 != 0)
    {
        // Extracting the last digit
        $d = (int)$n % 10;

        // Truncating the number
        $n /= 10;

        // Subtracting the five times
        // the last digit from the
        // remaining number
        $n -= $d * 5;
    }

    // Return n is divisible by 17
    return ($n % 17 == 0);
}

// Driver code
$n = 19877658;
if (isDivisible($n))
    print("Yes");
else
    print("No");

// This code is contributed by Raj
?>

Javascript

<script>

// JavaScript Program to validate the above logic

    // Function to check if the
// number is divisible by 17 or not
    function isDivisible(n)
    {
        while (Math.floor(n / 100)>0)
    {
        // Extracting the last digit
        let d = n % 10;

        // Truncating the number
        n = Math.floor(n/10);

        // Subtracting the five times the
        // last digit from the remaining number
        n -= d * 5;
    }

    // Return n is divisible by 17
    return (n % 17 == 0);
    }

    // Driver code
    let n = 19877658;
    if (isDivisible(n))
        document.write( "Yes");
    else
        document.write( "No");



// This code is contributed by avanitrachhadiya2155

</script>

Output:

Yes

Time Complexity: O(log₁₀n), time required to check if number is divisible by 17
Auxiliary Space: O(1), as no extra space is required

Note that the above program may not make a lot of sense as could simply do n % 23 to check for divisibility. The idea of this program is to validate the concept. Also, this might be an efficient approach if input number is large and given as string.