Check if a large number is divisible by a number which is a power of 2

Last Updated : 11 May, 2021

Given a large number in the form of a string str and a number K, the task is to check if the number formed by string str is divisible by the K or not, where K is a power of 2.
Examples:

Input: str = "5426987513245621541524288", num = 64
Output: Yes
Explanation:
Since log₂(64) = 6, so the number formed by the last 6 digits from the string str is divisible by 64 .
Input: str = "21346775656413259795656497974113461254", num = 4
Output: No
Explanation:
Since log₂(4)=2, the number formed by the last 2 digits from the string str is not divisible by 4.

Approach:
Since K is a perfect power of 2. Let K can be represented as 2^X. Then according to the divisibility rule of perfect power of 2, if the last X digit of the given number is divisible by K then the given number is divisible by K. Otherwise it is not divisible by K.
Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h> 
using namespace std;

// Function to check divisibility
bool checkIfDivisible(string str,
                      long long int num)
{

    // Calculate the number of digits in num
    long long int powerOf2 = log2(num);

    // Check if the length of
    // the string is less than
    // the powerOf2 then
    // return false
    if (str.length() < powerOf2)
        return false;

    // Check if the powerOf2 is 0
    // that means the given number
    // is 1 and as every number
    // is divisible by 1 so return true
    if (powerOf2 == 0)
        return true;

    // Find the number which is
    // formed by the last n digits
    // of the string where n=powerOf2
    long long int i, number = 0;
    int len = str.length();

    for (i = len - powerOf2; i < len; i++) {
        number += (str[i] - '0')
                  * pow(10,
                        powerOf2 - 1);
        powerOf2--;
    }

    // Check if the number formed is
    // divisible by input num or not
    if (number % num)
        return false;
    else
        return true;
}

// Driver Code
int main()
{
    // Given number
    string str = "213467756564";
    long long int num = 4;

    // Function Call
    if (checkIfDivisible(str, num))
        cout << "Yes";
    else
        cout << "No";

    return 0;
}

Java

// Java program for the above approach
class GFG{ 

// Function to check divisibility
static boolean checkIfDivisible(String str,
                                long num)
{
    
    // Calculate the number of digits in num
    long powerOf2 = (int)(Math.log(num) /
                          Math.log(2));

    // Check if the length of
    // the string is less than
    // the powerOf2 then
    // return false
    if (str.length() < powerOf2)
        return false;

    // Check if the powerOf2 is 0
    // that means the given number
    // is 1 and as every number
    // is divisible by 1 so return true
    if (powerOf2 == 0)
        return true;

    // Find the number which is
    // formed by the last n digits
    // of the string where n=powerOf2
    long i, number = 0;
    int len = str.length();

    for(i = len - powerOf2; i < len; i++) 
    {
        number += (str.charAt((int)i) - '0') * 
                   Math.pow(10, powerOf2 - 1);
        powerOf2--;
    }

    // Check if the number formed is
    // divisible by input num or not
    if (number % num != 0)
        return false;
    else
        return true;
}

// Driver Code
public static void main(String[] args) 
{
    
    // Given number
    String str = "213467756564";
    long num = 4;
    
    // Function call
    if (checkIfDivisible(str, num))
        System.out.print("Yes");
    else
        System.out.print("No");
}
}

// This code is contributed by rutvik_56

Python3

# Python3 program for the above approach 
from math import log2

# Function to check divisibility 
def checkIfDivisible(string, num): 

    # Calculate the number of digits in num 
    powerOf2 = int(log2(num)); 

    # Check if the length of 
    # the string is less than 
    # the powerOf2 then 
    # return false 
    if (len(string) < powerOf2):
        return False; 

    # Check if the powerOf2 is 0 
    # that means the given number 
    # is 1 and as every number 
    # is divisible by 1 so return true 
    if (powerOf2 == 0):
        return True; 

    # Find the number which is 
    # formed by the last n digits 
    # of the string where n=powerOf2 
    number = 0; 
    length = len(string); 

    for i in range(length - powerOf2, length): 
        number += ((ord(string[i]) - ord('0')) *
                  (10 ** (powerOf2 - 1))); 
        
        powerOf2 -= 1; 

    # Check if the number formed is 
    # divisible by input num or not 
    if (number % num):
        return False; 
    else :
        return True; 

# Driver Code 
if __name__ == "__main__" : 

    # Given number 
    string = "213467756564"; 
    num = 4; 

    # Function Call 
    if (checkIfDivisible(string, num)):
        print("Yes"); 
    else :
        print("No"); 

# This code is contributed by AnkitRai01

// C# program for the above approach
using System;

class GFG{ 

// Function to check divisibility
static bool checkIfDivisible(String str,
                             long num)
{
    
    // Calculate the number of digits in num
    long powerOf2 = (int)(Math.Log(num) /
                          Math.Log(2));

    // Check if the length of
    // the string is less than
    // the powerOf2 then
    // return false
    if (str.Length < powerOf2)
        return false;

    // Check if the powerOf2 is 0
    // that means the given number
    // is 1 and as every number
    // is divisible by 1 so return true
    if (powerOf2 == 0)
        return true;

    // Find the number which is
    // formed by the last n digits
    // of the string where n=powerOf2
    long i, number = 0;
    int len = str.Length;

    for(i = len - powerOf2; i < len; i++) 
    {
        number += (long)((str[(int)i] - '0') * 
                Math.Pow(10, powerOf2 - 1));
        powerOf2--;
    }

    // Check if the number formed is
    // divisible by input num or not
    if (number % num != 0)
        return false;
    else
        return true;
}

// Driver Code
public static void Main(String[] args) 
{
    
    // Given number
    String str = "213467756564";
    long num = 4;
    
    // Function call
    if (checkIfDivisible(str, num))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}

// This code is contributed by amal kumar choubey

JavaScript

<script>

// Javascript program for the above approach

// Function to check divisibility
function checkIfDivisible(str, num)
{
    
    // Calculate the number of digits in num
    let powerOf2 = (Math.log(num) /
                          Math.log(2));

    // Check if the length of
    // the string is less than
    // the powerOf2 then
    // return false
    if (str.length < powerOf2)
        return false;

    // Check if the powerOf2 is 0
    // that means the given number
    // is 1 and as every number
    // is divisible by 1 so return true
    if (powerOf2 == 0)
        return true;

    // Find the number which is
    // formed by the last n digits
    // of the string where n=powerOf2
    let i, number = 0;
    let len = str.length;

    for(i = len - powerOf2; i < len; i++) 
    {
        number += (str[i] - '0') * 
                   Math.pow(10, powerOf2 - 1);
        powerOf2--;
    }

    // Check if the number formed is
    // divisible by input num or not
    if (number % num != 0)
        return false;
    else
        return true;
}

// Driver Code
    
    // Given number
    let str = "213467756564";
    let num = 4;
    
    // Function call
    if (checkIfDivisible(str, num))
        document.write("Yes");
    else
        document.write("No");
  
</script>

Output:

Yes

Time Complexity: O(Len), where Len is the length of the string.
Auxiliary Space: O(log₂K)

Check if a large number is divisible by a number which is a power of 2

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Check if a large number is divisible by a number which is a power of 2

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