# Divisible by 37 for large numbers

Given a large number n, we need to check whether it is divisible by 37. Print true if it is divisible by 37 otherwise False.

Examples:

Input  : 74
Output : True

Input : 73
Output : False

Input : 8955795758 (10 digit number)
Output : True


## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A r digit number m whose digital form is (ar-1 ar-2….a2 a1 a0) is divisible by 37 if and only if the sum of series of numbers (a2 a1 a0) + (a5 a4 a3) + (a8 a7 a6) + … is divisible by 37. 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 37), 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 37),
Thus n is divisible by 37 if and if only if the series is divisible by 37.

Examples:

Input : 8955795758 (10 digit number)
Output : True
Explanation:
We express the number in terms of
triplets of digits as follows.
(008)(955)(795)(758)
Now, 758 + 795 + 955 + 8 = 2516
For 2516, the triplets will be:
(002)(516)
Now 516 + 2 = 518 which is divisible
by 37. Hence the number is divisible
by 37.

Input : 189710809179199 (15 digit number)
Output : False


## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 series . Calculate the sum of the series. If the sum of series has more than 3 digits in it, again recursively call this function.
Finally check whether the resultant sum is divisible by 37 or not.

Here is the program implementation to check divisibility by 37.

## C++

 // CPP program for checking divisibility by 37  // function divisible37 which returns True if   // number is divisible by 37 otherwise False  #include  using namespace std;     int divisibleby37(string n){      int l = n.length();      if (n == "0")          return 0;         // Append required 0's at the beginning      if (l % 3 == 1){          n = "00"+ n;          l += 2;      }      else if (l % 3 == 2){          n = "0"+ n;          l += 1;      }             int gSum = 0;             while (l != 0){         // group saves 3-digit group      string group = n.substr(l - 3, l);          l = l - 3;      int gvalue = (group - '0') * 100 +                    (group - '0') * 10 +                   (group - '0') * 1;                          // add the series      gSum = gSum + gvalue;      }             // if sum of series gSum has minimum 4       // digits in it, then again recursive       // call divisibleby37 function      if (gSum >= 1000)          return (divisibleby37(to_string(gSum)));      else         return (gSum % 37 == 0);     }     // drive program to test the above function  int main(){         string s="8955795758";             if (divisibleby37(s))      cout<<"True";      else     cout<<"False";      return 0;  }  // This code is contributed by Prerna Saini

## Java

 // Java program for checking   // divisibility by 37      class GFG   {   // function divisible37 which   // returns True if number is   // divisible by 37 otherwise False   static int divisibleby37(String n1)   {       int l = n1.length();      if (n1 == "0")           return 0;          // Append required 0's       // at the beginning       if (l % 3 == 1)       {           n1 = "00"+ n1;           l += 2;       }       else if (l % 3 == 2)       {           n1 = "0"+ n1;           l += 1;       }       char[]  n= n1.toCharArray();      int gSum = 0;       while (l != 0)       {          // group saves 3-digit group       int gvalue;       if(l == 2)           gvalue = ((int)n[(l - 2)] - 48) * 100 +                   ((int)n[(l - 1)] - 48) * 10;       else if(l == 1)           gvalue = ((int)n[(l - 1)] - 48) * 100;       else         gvalue = ((int)n[(l - 3)] - 48) * 100 +                   ((int)n[(l - 2)] - 48) * 10 +                   ((int)n[(l - 1)] - 48) * 1;       l = l - 3;              // add the series       gSum = gSum + gvalue;       }              // if sum of series gSum has minimum 4       // digits in it, then again recursive       // call divisibleby37 function       if (gSum >= 1000)           return (divisibleby37(String.valueOf(gSum)));       else         return (gSum % 37 == 0) ? 1 : 0;      }      // Driver Code   public static void main(String[] args)   {       String s="8955795758";              if (divisibleby37(s) == 1)       System.out.println("True");       else     System.out.println("False");   }   }      // This code is contributed by mits

## Python3

 # Python code for checking divisibility by 37  # function divisible37 which returns True if   # number is divisible by 37 otherwise False  def divisibleby37(n):      l = len(n)      if (n == 0):          return True          # Append required 0's at the beginning      if (l%3 == 1):          n = "00"+ n          l += 2     elif (l%3 == 2):          n = "0"+ n          l += 1        gSum = 0     while (l != 0):             # group saves 3-digit group          group = int(n[l-3:l])          l = l-3            # add the series          gSum = gSum + group         # if sum of series gSum has minimum 4       # digits in it, then again recursive       # call divisibleby37 function      if (gSum >= 1000):          return(divisibleby37(str(gSum)))      else:          return (gSum%37==0)     # Driver method to test the above function  print(divisibleby37("8955795758"))

## C#

 // C# program for checking   // divisibility by 37  using System;     class GFG  {  // function divisible37 which   // returns True if number is   // divisible by 37 otherwise False  static int divisibleby37(string n)  {      int l = n.Length;      if (n == "0")          return 0;         // Append required 0's      // at the beginning      if (l % 3 == 1)      {          n = "00"+ n;          l += 2;      }      else if (l % 3 == 2)      {          n = "0"+ n;          l += 1;      }             int gSum = 0;      while (l != 0)      {         // group saves 3-digit group      int gvalue;      if(l == 2)          gvalue = ((int)n[(l - 2)] - 48) * 100 +                    ((int)n[(l - 1)] - 48) * 10;      else if(l == 1)          gvalue = ((int)n[(l - 1)] - 48) * 100;      else         gvalue = ((int)n[(l - 3)] - 48) * 100 +                    ((int)n[(l - 2)] - 48) * 10 +                   ((int)n[(l - 1)] - 48) * 1;      l = l - 3;             // add the series      gSum = gSum + gvalue;      }             // if sum of series gSum has minimum 4       // digits in it, then again recursive       // call divisibleby37 function      if (gSum >= 1000)          return (divisibleby37(gSum.ToString()));      else         return (gSum % 37 == 0) ? 1 : 0;     }     // Driver Code  public static void Main()  {      string s="8955795758";             if (divisibleby37(s) == 1)      Console.WriteLine("True");      else     Console.WriteLine("False");  }  }     // This code is contributed by mits

## PHP

 = 1000)          return (divisibleby37((string)($gSum)));   else  return ($gSum % 37 == 0);     }     // Driver code  $s = "8955795758";    if (divisibleby37($s))  echo "True";  else echo "False";     // This code is contributed  // by mits  ?>

Output:

True


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