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 last digit from 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
Substract 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 $\overline{a b c}$ be any number such that $\overline{a b c}$ =100a+10b+c .
Now assume that $\overline{a b c}$ is divisible by 17. Then
$\overline{a b c}\equiv$ 0 (mod 17)
100a+10b+c $\equiv$ 0 (mod 17)
10(10a+b)+c $\equiv$ 0 (mod 17)
10 $\overline{a b}$ +c $\equiv$ 0 (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 $\overline{a b}$ 1.
In other words, we have to find an integer such that n such that 10n $\equiv$ 1 mod 17.
It can be observed that the smallest n which satisfies this property is -5 as -50 $\equiv$ 1 mod 17.
Now we can multiply the original equation 10 $\overline{a b}$ +c $\equiv$ 0 (mod 17)
by -5 and simplify it:
-50 $\overline{a b}$ -5c $\equiv$ 0 (mod 17)
$\overline{a b}$ -5c $\equiv$ 0 (mod 17)
We have found out that if $\overline{a b c}\equiv$ 0 (mod 17) then,
$\overline{a b}$ -5c $\equiv$ 0 (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 
?> 

Output:

Yes

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.

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My Personal Notes

C++

Java

Python 3

C#

PHP

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