Given a number, check whether a given number is divisible by 36 or not. The number may be very large and may not fit in any numeric(int, long int, float, etc.) data type.
Examples:
Input : 72
Output : Yes
Input : 244
Output : No
Input : 11322134
Output : No
Input : 92567812197966231384
Output : Yes
A number is divisible by 36 if the number is divisible by 4 and 9
- A number is divisible by 4 if the number formed by its last 2 digits is divisible by 4
- A number is divisible by 9 if the sum of the digits of the number is divisible by 9
Below is the implementation based on the above idea.
C++
// C++ implementation to check divisibility by 36#include <bits/stdc++.h>using namespace std;// Function to check whether a number// is divisible by 36 or notstring divisibleBy36(string num){ int l = num.length(); // null number cannot // be divisible by 36 if (l == 0) return "No"; // single digit number other than // 0 is not divisible by 36 if (l == 1 && num[0] != '0') return "No"; // number formed by the last 2 digits int two_digit_num = (num[l-2] - '0')*10 + (num[l-1] - '0') ; // if number is not divisible by 4 if (two_digit_num%4 != 0) return "No"; // number is divisible by 4 calculate // sum of digits int sum = 0; for (int i=0; i<l; i++) sum += (num[i] - '0'); // sum of digits is not divisible by 9 if (sum%9 != 0) return "No"; // number is divisible by 4 and 9 // hence, number is divisible by 36 return "Yes";}// Driver programint main(){ string num = "92567812197966231384"; cout << divisibleBy36(num); return 0;} |
Java
// Java program to find if a number is// divisible by 36 or notimport java.io.*;class IsDivisible{ // Function to check whether a number // is divisible by 36 or not static boolean divisibleBy36(String num) { int l = num.length(); // null number cannot // be divisible by 36 if (l == 0) return false; // single digit number other than // 0 is not divisible by 36 if (l == 1 && num.charAt(0) != '0') return false; // number formed by the last 2 digits int two_digit_num = (num.charAt(l-2) - '0')*10 + (num.charAt(l-1) - '0') ; // if number is not divisible by 4 if (two_digit_num%4 != 0) return false; // number is divisible by 4 calculate // sum of digits int sum = 0; for (int i=0; i<l; i++) sum += (num.charAt(i) - '0'); // sum of digits is not divisible by 9 if (sum%9 != 0) return false; // number is divisible by 4 and 9 // hence, number is divisible by 36 return true; } // main function public static void main (String[] args) { String num = "92567812197966231384"; if(divisibleBy36(num)) System.out.println("Yes"); else System.out.println("No"); }} |
Python3
# Python 3 implementation to# check divisibility by 36# Function to check whether a# number is divisible by # 36 or notdef divisibleBy36(num) : l = len(num) # null number cannot # be divisible by 36 if (l == 0) : return ("No") # single digit number other # than 0 is not divisible # by 36 if (l == 1 and num[0] != '0') : return ("No") # number formed by the # last 2 digits two_digit_num = (((int)(num[l - 2])) * 10 +(int)(num[l - 1])) # if number is not # divisible by 4 if (two_digit_num%4 != 0) : return "No" # number is divisible # by 4 calculate sum # of digits sm = 0 for i in range(0,l) : sm = sm + (int)(num[i]) # sum of digits is not # divisible by 9 if (sm%9 != 0) : return ("No") # Number is divisible # by 4 and 9 hence, # number is divisible # by 36 return ("Yes")# Driver programnum = "92567812197966231384"print(divisibleBy36(num))# This code is contributed by Nikita Tiwari. |
C#
// C# program to find if a number is// divisible by 36 or notusing System;class GFG { // Function to check whether // a number is divisible by // 36 or not static bool divisibleBy36(String num) { int l = num.Length; // null number cannot // be divisible by 36 if (l == 0) return false; // single digit number other than // 0 is not divisible by 36 if (l == 1 && num[0] != '0') return false; // number formed by the last // 2 digits int two_digit_num = (num[l-2] - '0') * 10 + (num[l-1] - '0') ; // if number is not divisible by 4 if (two_digit_num % 4 != 0) return false; // number is divisible by 4 calculate // sum of digits int sum = 0; for (int i = 0; i < l; i++) sum += (num[i] - '0'); // sum of digits is not divisible by 9 if (sum % 9 != 0) return false; // number is divisible by 4 and 9 // hence, number is divisible by 36 return true; } // main function public static void Main () { String num = "92567812197966231384"; if(divisibleBy36(num)) Console.Write("Yes"); else Console.Write("No"); }} // This code is contributed by parashar. |
Javascript
<script>// Javascript implementation to// check divisibility by 36// Function to check whether a number// is divisible by 36 or notfunction divisibleBy36(num){ let l = num.length; // null number cannot // be divisible by 36 if (l == 0) return "No"; // single digit number other than // 0 is not divisible by 36 if (l == 1 && num[0] != '0') return "No"; // number formed by the // last 2 digits let two_digit_num = (num[l - 2] - '0') * 10 + (num[l - 1] - '0') ; // if number is not // divisible by 4 if (two_digit_num%4 != 0) return "No"; // number is divisible by 4 // calculate sum of digits let sum = 0; for (let i = 0; i < l; i++) sum += (num[i] - '0'); // sum of digits is not // divisible by 9 if (sum % 9 != 0) return "No"; // number is divisible by 4 and 9 // hence, number is divisible by 36 return "Yes";}// Driver Codelet num = "92567812197966231384";document.write(divisibleBy36(num));// This code is contributed by _saurabh_jaiswal.</script> |
PHP
<?php// PHP implementation to // check divisibility by 36// Function to check whether a number// is divisible by 36 or notfunction divisibleBy36($num){ $l = strlen($num); // null number cannot // be divisible by 36 if ($l == 0) return "No"; // single digit number other than // 0 is not divisible by 36 if ($l == 1 && $num[0] != '0') return "No"; // number formed by the // last 2 digits $two_digit_num = ($num[$l - 2] - '0') * 10 + ($num[$l - 1] - '0') ; // if number is not // divisible by 4 if ($two_digit_num%4 != 0) return "No"; // number is divisible by 4 // calculate sum of digits $sum = 0; for ($i = 0; $i < $l; $i++) $sum += ($num[$i] - '0'); // sum of digits is not // divisible by 9 if ($sum % 9 != 0) return "No"; // number is divisible by 4 and 9 // hence, number is divisible by 36 return "Yes";}// Driver Code$num = "92567812197966231384";echo(divisibleBy36($num));// This code is contributed by Ajit.?> |
Output
Yes
Time Complexity: O(n)
Auxiliary Space: O(1) because constant space for variables is being used
Method 2: Checking given number is divisible by 36 or not by using the modulo division operator “%”.
C++
#include <iostream>using namespace std;int main(){ //input long long int n = 72; // finding given number is divisible by 36 or not if (n % 36 == 0) { cout << "Yes"; } else { cout << "No"; } return 0;}// This code is contributed by satwik4409. |
Java
/*package whatever //do not write package name here */import java.io.*;class GFG { public static void main (String[] args) { // Java code for the above approach // To check whether the given number is divisible by 36 or not //input long n = 72; // finding given number is divisible by 36 or not if (n % 36 == 0) System.out.println("Yes"); else System.out.println("No"); }}// This code is contributed by laxmigangarajula03 |
Python3
# Python code # To check whether the given number is divisible by 36 or not#input n=72# the above input can also be given as n=input() -> taking input from user# finding given number is divisible by 36 or notif int(n)%36==0: print("Yes") else: print("No") # this code is contributed by gangarajula laxmi |
C#
using System;public class GFG{ static public void Main () { // C# code for the above approach // To check whether the given number is divisible by 36 or not //input long n = 72; // finding given number is divisible by 36 or not if (n % 36 == 0) Console.Write("Yes"); else Console.Write("No"); } }// this code is contributed by ksrikanth0498 |
Javascript
<script> // JavaScript code for the above approach // To check whether the given number is divisible by 36 or not //input n = 72 // finding given number is divisible by 36 or not if (n % 36 == 0) document.write("Yes") else document.write("No") // This code is contributed by Potta Lokesh </script> |
PHP
<?php $num = 72; // checking if the given number is divisible by 36 or // not using modulo division operator if the output of // num%36 is equal to 0 then given number is divisible // by 36 otherwise not divisible by 36 if ($num % 36 == 0) { echo "Yes"; } else { echo "No"; } ?> |
Output
Yes
Time complexity: O(1) as it is doing constant operations
Auxiliary space: O(1)
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Approach 3:
Approach:
We can convert the number to a string and use string slicing to check if the last two digits are divisible by 4. Then we can loop through the digits of the number using a for loop and calculate their sum using the built-in sum() function. Finally, we can check if the digit sum is divisible by 9.
- Check if the last two digits of the number are divisible by 4 using the modulo operator.
- Convert the number to a string and loop through its digits using a for loop.
- Convert each digit to an integer using the int() function and calculate their sum using the built-in sum() function.
- Check if the digit sum is divisible by 9 using the modulo operator.
- Return True if both conditions are met, else False.
C++
#include <iostream>#include <string>using namespace std;bool isDivisibleBy36(int n) { // Check if the last two digits of the number are divisible by 4 if (stoi(to_string(n).substr(to_string(n).length() - 2)) % 4 != 0) { return false; } int digitSum = 0; string numStr = to_string(n); // Calculate the sum of all the digits in the number for (char digit : numStr) { digitSum += digit - '0'; // Convert char to int and add to the sum } // Check if the sum of digits is divisible by 9 return digitSum % 9 == 0;}int main() { long long n = 92567812197966231384; // Function Call if (isDivisibleBy36(n)) { cout << "Yes" << endl; } else { cout << "No" << endl; } return 0;} |
Java
import java.math.BigInteger;public class Main { // Function to check if a number is divisible by 36 public static boolean isDivisibleBy36(BigInteger n) { // Check if the last two digits of the number are divisible by 4 if (!n.remainder(BigInteger.valueOf(4)).equals(BigInteger.ZERO)) { return false; } BigInteger digitSum = BigInteger.ZERO; String numStr = n.toString(); // Calculate the sum of all the digits in the number for (char digit : numStr.toCharArray()) { digitSum = digitSum.add(BigInteger.valueOf(digit - '0')); } // Check if the sum of digits is divisible by 9 return digitSum.remainder(BigInteger.valueOf(9)).equals(BigInteger.ZERO); } public static void main(String[] args) { BigInteger n = new BigInteger("92567812197966231384"); // Function Call if (isDivisibleBy36(n)) { System.out.println("Yes"); } else { System.out.println("No"); } }} |
Python3
import timedef is_divisible_by_36_2(n): if int(str(n)[-2:]) % 4 != 0: return False digit_sum = sum(int(digit) for digit in str(n)) return digit_sum % 9 == 0print(is_divisible_by_36_2(92567812197966231384)) |
Javascript
function isDivisibleBy36(n) { // Check if the last two digits of n are divisible by 4 if (parseInt(n.toString().slice(-2)) % 4 !== 0) { return false; } // Convert n to a string and split it into an array of digits, // then convert each digit back to a number const digitSum = n .toString() .split('') .map(Number) // Calculate the sum of all the digits in the number .reduce((acc, digit) => acc + digit, 0); // Check if the sum of digits is divisible by 9 return digitSum % 9 === 0;}// Test the function with a large numberconsole.log(isDivisibleBy36(BigInt("92567812197966231384"))); // Should output true |
Output
True
The time complexity of this approach is also O(log(n)), as we loop through the digits of the number.
The space complexity is O(log(n)), as we convert the number to a string.
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