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Check if any large number is divisible by 17 or not
• Difficulty Level : Easy
• Last Updated : 25 Oct, 2018

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 be any number such that =100a+10b+c .
Now assume that is divisible by 17. Then
0 (mod 17)
100a+10b+c 0 (mod 17)
10(10a+b)+c 0 (mod 17)
10+c 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 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+c 0 (mod 17)
by -5 and simplify it:
-50-5c 0 (mod 17)
-5c 0 (mod 17)
We have found out that if 0 (mod 17) then,
-5c 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   using namespace std;  // Function to check if the// number is divisible by 17 or notbool 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 codeint 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 Codeif __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

 
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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