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: YesInput: N = 54
Output: No
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 notbool 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 Codeint 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 notimport java.util.*; class GFG{ // Function to check if the// number is divisible by 53 or notstatic 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 Codepublic 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 notdef 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 Codeif __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 notstatic 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 Codepublic 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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