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

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

{\displaystyle a\equiv b{\pmod {m}}\,}

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

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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// CPP program for checking divisibility by 37
// function divisible37 which returns True if 
// number is divisible by 37 otherwise False
#include <bits/stdc++.h>
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] - '0') * 100 + 
                 (group[1] - '0') * 10 +
                 (group[2] - '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

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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 

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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"))

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP program for checking 
// divisibility by 37
  
// function divisible37 which 
// returns True if number is 
// divisible by 37 otherwise
// False
function divisibleby37($n)
{
    $l = strlen($n);
    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;
    }
      
    $gSum = 0;
      
    while ($l != 0)
    {
  
    // group saves 3-digit group
    $group = substr($n,$l - 3, $l);
        $l = $l - 3;
    $gvalue = (ord($group[0]) - 48) * 100 + 
              (ord($group[1]) - 48) * 10 +
              (ord($group[2]) - 48) * 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 ($gSum >= 1000)
        return (divisibleby37((string)($gSum)));
    else
        return ($gSum % 37 == 0);
  
}
  
// Driver code
$s = "8955795758";
  
if (divisibleby37($s))
echo "True";
else
echo "False";
  
// This code is contributed
// by mits
?>

chevron_right



Output:

True

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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



My Personal Notes arrow_drop_up

Improved By : Mithun Kumar



Article Tags :
Practice Tags :


Be the First to upvote.


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