# To check whether a large number is divisible by 7

You are given an n-digit large number, you have to check whether it is divisible by 7.
A (r+1)-digit integer n whose digital form is (ar ar-1 ar-2….a2 a1 a0) is divisible by 7 if and only if the alternate series of numbers (a2 a1 a0) – (a5 a4 a3) + (a8 a7 a6) – … is divisible by 7.
The triplets of digits within parenthesis represent 3-digit number in digital form.

The given number n can be written as a sum of powers of 1000 as follows.
n= (a2 a1 a0) + (a5 a4 a3)*1000 + (a8 a7 a6)*(1000*1000) +….
As 1000 = (-1)(mod 7), 1000 as per congruence relation.
For a positive integer n, two numbers a and b are said to be congruent modulo n, if their difference
(a – b) is an integer multiple of n (that is, if there is an integer k such that a – b = kn). This congruence relation is typically considered when a and b are integers, and is denoted Hence we can write:
n = { (a2a1a0) + (a5a4a3)* (-1) + (a8a7a6)* (-1)*(-1)+…..}(mod 7),
Thus n is divisible by 7 if and if only if the series is divisible by 7.

Examples :

Input : 8955795758
Output : Divisible by 7
Explanation:
We express the number in terms of triplets
of digits as follows.
(008)(955)(795)(758)
Now, 758- 795 + 955 - 8 = 910, which is
divisible by 7

Input : 100000000000
Output : Not Divisible by 7
Explanation:
We express the number in terms of triplets
of digits as follows.
(100)(000)(000)(000)
Now, 000- 000 + 000 - 100 = -100, which is
not divisible by 7


Note that the number of digits in n may not be multiple of 3 . In that case we pas zero(s) on the left side of the remaining digits(s) after taking out all the triplets (from right side of n) to form the last triplet.
A simple and efficient method is to take input in form of string (make its length in form of 3*m by adding 0 to left of number if required) and then you have to add the digits in blocks of three from right to left until it become a 3 digit number to form an alternate series and check whether the series is divisible by 7 or not.
Here the program implementation to check divisibility of 7 is done.

## C++

 // C++ code to check divisibility of a  // given large number by 7  #include  using namespace std;     int isdivisible7(char num[])  {      int n = strlen(num), gSum;      if (n == 0 && num == '\n')          return 1;         // Append required 0s at the beginning.      if (n % 3 == 1) {          strcat(num, "00");          n += 2;      }      else if (n % 3 == 2) {          strcat(num, "0");          n++;      }         // add digits in group of three in gSum      int i, GSum = 0, p = 1;      for (i = n - 1; i >= 0; i--) {             // group saves 3-digit group          int group = 0;          group += num[i--] - '0';          group += (num[i--] - '0') * 10;          group += (num[i] - '0') * 100;             gSum = gSum + group * p;             // generate alternate series of plus          // and minus          p *= (-1);      }         return (gSum % 7 == 0);  }     // Driver code  int main()  {      // Driver method      char num[] = "8955795758";      if (isdivisible7(num))          cout << "Divisible by 7";      else         cout << "Not Divisible by 7";      return 0;  }     // This code is contributed  // by Akanksha Rai

## C

 // C code to check divisibility of a  // given large number by 7  #include  #include  int isdivisible7(char num[])  {      int n = strlen(num), gSum;      if (n == 0 && num == '\n')          return 1;         // Append required 0s at the beginning.      if (n % 3 == 1) {          strcat(num, "00");          n += 2;      }      else if (n % 3 == 2) {          strcat(num, "0");          n++;      }         // add digits in group of three in gSum      int i, GSum = 0, p = 1;      for (i = n - 1; i >= 0; i--) {             // group saves 3-digit group          int group = 0;          group += num[i--] - '0';          group += (num[i--] - '0') * 10;          group += (num[i] - '0') * 100;             gSum = gSum + group * p;             // generate alternate series of plus          // and minus          p *= (-1);      }         return (gSum % 7 == 0);  }     // Driver code  int main()  {      // Driver method      char num[] = "8955795758";      if (isdivisible7(num))          printf("Divisible by 7");      else         printf("Not Divisible by 7");      return 0;  }

## Java

 // Java code to check divisibility of a given large number by 7     class Test {      // Method to check divisibility      static boolean isDivisible7(String num)      {          int n = num.length();          if (n == 0 && num.charAt(0) == '0')              return true;             // Append required 0s at the beginning.          if (n % 3 == 1)              num = "00" + num;          if (n % 3 == 2)              num = "0" + num;          n = num.length();             // add digits in group of three in gSum          int gSum = 0, p = 1;          for (int i = n - 1; i >= 0; i--) {                 // group saves 3-digit group              int group = 0;              group += num.charAt(i--) - '0';              group += (num.charAt(i--) - '0') * 10;              group += (num.charAt(i) - '0') * 100;              gSum = gSum + group * p;              // generate alternate series of plus and minus              p = p * -1;          }             // calculate result till 3 digit sum          return (gSum % 7 == 0);      }         // Driver method      public static void main(String args[])      {          String num = "8955795758";             System.out.println(isDivisible7(num) ? "Divisible by 7" : "Not Divisible  by 7");      }  }

## Python 3

 # Python 3 code to check divisibility   # of a given large number by 7     def isdivisible7(num):      n = len(num)      if (n == 0 and num == '\n'):          return 1        # Append required 0s at the beginning.      if (n % 3 == 1) :          num = str(num) + "00"         n += 2            elif (n % 3 == 2) :          num = str(num) + "0"         n += 1        # add digits in group of three in gSum      GSum = 0     p = 1     for i in range(n - 1, -1, -1) :             # group saves 3-digit group          group = 0         group += ord(num[i]) - ord('0')          i -= 1         group += (ord(num[i]) - ord('0')) * 10         i -= 1         group += (ord(num[i]) - ord('0')) * 100            GSum = GSum + group * p             # generate alternate series of           # plus and minus          p *= (-1)         return (GSum % 7 == 0)     # Driver code  if __name__ == "__main__":             num = "8955795758"     if (isdivisible7(num)):          print("Divisible by 7")      else :          print("Not Divisible by 7")     # This code is contributed by ChitraNayal

## C#

 // C# code to check divisibility of a  // given large number by 7  using System;     class GFG {         // Method to check divisibility      static bool isDivisible7(String num)      {          int n = num.Length;          if (n == 0 && num == '0')              return true;             // Append required 0s at the beginning.          if (n % 3 == 1)              num = "00" + num;             if (n % 3 == 2)              num = "0" + num;             n = num.Length;             // add digits in group of three in gSum          int gSum = 0, p = 1;          for (int i = n - 1; i >= 0; i--) {                 // group saves 3-digit group              int group = 0;              group += num[i--] - '0';              group += (num[i--] - '0') * 10;              group += (num[i] - '0') * 100;              gSum = gSum + group * p;                 // generate alternate series              // of plus and minus              p = p * -1;          }             // calculate result till 3 digit sum          return (gSum % 7 == 0);      }         // Driver code      static public void Main()      {          String num = "8955795758";             // Function calling          Console.WriteLine(isDivisible7(num) ? "Divisible by 7" : "Not Divisible by 7");      }  }     // This code is contributed by Ajit.

## PHP

 = 0; $i--)   {     // group saves 3-digit group   $group = 0;           $group += $num[$i--] - '0';   $group += ($num[$i--] - '0') * 10;           $group += ($num[$i] - '0') * 100;   $gSum = $gSum + $group * $p;     // generate alternate series   // of plus and minus   $p = $p * -1;   }     // calculate result till 3 digit sum   return ($gSum % 7 == 0);   }      // Driver Code  $num = "8955795758";    echo (isDivisible7($num) ?           "Divisible by 7" :          "Not Divisible by 7");      // This code is contributed by Ryuga  ?>

Output:

 Divisible by 7


This article is contributed by Sruti Rai.If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up

Article Tags :
Practice Tags :

1

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.