Check whether a large number is divisible by 53 or not

Given a large number in the form of a string N, the task is to check whether the number is divisible by 53 or not.

Examples:

Input: N = 5299947
Output: Yes

Input: N = 54
Output: No

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

Approach:

Extract the last digit of the given string N and remove it.
Multiply that digit by 37.
Subtract the product calculated in the above step from the remaining number.
Continue until we reduce the given string to a 3 or four digit number.
Convert the remaining string to its corresponding integer form and check if it is divisible by 53 or not.

Below is the implementation of the above approach:

C++

// C++ program to check 
// whether a number 
// is divisible by 53 or not 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to check if the 
// number is divisible by 53 or not 
bool isDivisible(string s) 
{ 
    int flag = 0; 
    while (s.size() > 4) { 
  
        int l = s.size() - 1; 
        int x = (s[l] - '0') * 37; 
  
        reverse(s.begin(), s.end()); 
        s.erase(0, 1); 
  
        int i = 0, carry = 0; 
        while (x) { 
            int d = (s[i] - '0') 
                    - (x % 10) 
                    - carry; 
            if (d < 0) { 
                d += 10; 
                carry = 1; 
            } 
            else
                carry = 0; 
  
            s[i] = (char)(d + '0'); 
            x /= 10; 
            i++; 
        } 
  
        while (carry && i < l) { 
            int d = (s[i] - '0') - carry; 
            if (d < 0) { 
                d += 10; 
                carry = 1; 
            } 
            else
                carry = 0; 
  
            s[i] = (char)(d + '0'); 
  
            i++; 
        } 
  
        reverse(s.begin(), s.end()); 
    } 
  
    int num = 0; 
    for (int i = 0; i < s.size(); i++) { 
        num = num * 10 + (s[i] - '0'); 
    } 
  
    if (num % 53 == 0) 
        return true; 
    else
        return false; 
} 
  
// Driver Code 
int main() 
{ 
    string N = "18432462191076"; 
  
    if (isDivisible(N)) 
        cout << "Yes" << endl; 
    else
        cout << "No" << endl; 
  
    return 0; 
} 

Java

// Java program to check whether  
// a number is divisible by 53 or not 
import java.util.*; 
  
class GFG{ 
  
// Function to check if the 
// number is divisible by 53 or not 
static boolean isDivisible(char []s) 
{ 
    while (s.length > 4) 
    { 
        int l = s.length - 1; 
        int x = (s[l] - '0') * 37; 
  
        s = reverse(s); 
        s = Arrays.copyOfRange(s, 1, s.length); 
  
        int i = 0, carry = 0; 
        while (x > 0)  
        { 
            int d = (s[i] - '0') -  
                        (x % 10) -  
                        carry; 
                      
            if (d < 0) 
            { 
                d += 10; 
                carry = 1; 
            } 
            else
                carry = 0; 
  
            s[i] = (char)(d + '0'); 
            x /= 10; 
            i++; 
        } 
  
        while (carry > 0 && i < l) 
        { 
            int d = (s[i] - '0') - carry; 
            if (d < 0) 
            { 
                d += 10; 
                carry = 1; 
            } 
            else
                carry = 0; 
  
            s[i] = (char)(d + '0'); 
            i++; 
        } 
        s = reverse(s); 
    } 
  
    int num = 0; 
    for(int i = 0; i < s.length; i++)  
    { 
       num = num * 10 + (s[i] - '0'); 
    } 
  
    if (num % 53 == 0) 
        return true; 
    else
        return false; 
} 
  
static char[] reverse(char []a) 
{ 
    int l, r = a.length - 1; 
      
    for(l = 0; l < r; l++, r--) 
    { 
       char temp = a[l]; 
            a[l] = a[r]; 
            a[r] = temp; 
    } 
    return a; 
} 
  
// Driver Code 
public static void main(String[] args) 
{ 
    String N = "18432462191076"; 
  
    if (isDivisible(N.toCharArray())) 
        System.out.print("Yes" + "\n"); 
    else
        System.out.print("No" + "\n"); 
} 
} 
  
// This code is contributed by Rohit_ranjan 

Python3

# Python3 program to check whether a   
# number is divisible by 53 or not 
  
# Function to check if the 
# number is divisible by 53 or not 
def isDivisible(s): 
      
    flag = 0
      
    while (len(s) > 4): 
        l = len(s) - 1
        x = (ord(s[l]) - ord('0')) * 37
  
        s = s[::-1] 
        s = s.replace('0', '', 1) 
  
        i = 0
        carry = 0
          
        while (x): 
            d = ((ord(s[i]) - ord('0')) - 
                 (x % 10) - carry) 
            if (d < 0): 
                d += 10
                carry = 1
            else: 
                carry = 0
                  
            s = s.replace(s[i], chr(d + ord('0')), 1) 
            x //= 10
            i += 1
  
        while (carry and i < l): 
            d = (ord(s[i]) - ord('0')) - carry 
              
            if (d < 0): 
                d += 10
                carry = 1
            else: 
                carry = 0
              
            s = s.replace(s[i], chr(d + ord('0')), 1) 
            i += 1
        s = s[::-1] 
  
    num = 0
    for i in range(len(s)): 
        num = num * 10 + (ord(s[i]) - ord('0')) 
  
    if (num % 53 == 0): 
        return True
    else: 
        return False
  
# Driver Code 
if __name__ == '__main__': 
      
    N = "1843246219106"
  
    if (isDivisible(N)): 
        print("No") 
    else: 
        print("Yes") 
  
# This code is contributed by Surendra_Gangwar 

C#

// C# program to check whether   
// a number is divisible by 53 or not  
using System;  
using System.Collections;  
using System.Collections.Generic;  
  
class GFG{ 
      
// Function to check if the 
// number is divisible by 53 or not 
static bool isDivisible(char []s) 
{ 
    while (s.Length > 4) 
    { 
        int l = s.Length - 1; 
        int x = (s[l] - '0') * 37; 
  
        s = reverse(s); 
          
        char []tmp = new char[s.Length - 1]; 
          
        Array.Copy(s, 1, tmp, 0, s.Length - 1); 
        s = tmp; 
          
        int i = 0, carry = 0; 
        while (x > 0)  
        { 
            int d = (s[i] - '0') -  
                       (x % 10) - carry; 
                         
            if (d < 0) 
            { 
                d += 10; 
                carry = 1; 
            } 
            else
                carry = 0; 
  
            s[i] = (char)(d + '0'); 
            x /= 10; 
            i++; 
        } 
  
        while (carry > 0 && i < l) 
        { 
            int d = (s[i] - '0') - carry; 
            if (d < 0) 
            { 
                d += 10; 
                carry = 1; 
            } 
            else
                carry = 0; 
  
            s[i] = (char)(d + '0'); 
            i++; 
        } 
        s = reverse(s); 
    } 
  
    int num = 0; 
    for(int i = 0; i < s.Length; i++)  
    { 
        num = num * 10 + (s[i] - '0'); 
    } 
  
    if (num % 53 == 0) 
        return true; 
    else
        return false; 
} 
  
static char[] reverse(char []a) 
{ 
    int l, r = a.Length - 1; 
      
    for(l = 0; l < r; l++, r--) 
    { 
        char temp = a[l]; 
             a[l] = a[r]; 
             a[r] = temp; 
    } 
    return a; 
} 
  
// Driver Code 
public static void Main(string[] args) 
{ 
    string N = "18432462191076"; 
  
    if (isDivisible(N.ToCharArray())) 
        Console.Write("Yes" + "\n"); 
    else
        Console.Write("No" + "\n"); 
} 
} 
  
// This code is contributed by rutvik_56 

Output:

Yes

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My Personal Notes

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

C++

Java

Python3

C#

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