# 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 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 10n 1 mod 17.
It can be observed that the smallest n which satisfies this property is -5 as -50 1 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 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

 

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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