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Number of digits in 2 raised to power n

  • Difficulty Level : Medium
  • Last Updated : 06 Apr, 2021

Let n be any power raised to base 2 i.e 2n. We are given the number n and our task is to find out the number of digits contained in the number 2n.
Examples: 

Input : n = 5
Output : 2
Explanation : 2n = 32, which has only
2 digits.

Input : n = 10
Output : 4
Explanation : 2n = 1024, which has only
4 digits.

We can write 2n using logarithms as:  

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2n = 10nlog102

Now suppose, x = nlog102, 
Therefore, 2n = 10x
Also, we all know that the number, 10n will have (n+1) digits. Therefore, 10x will have (x+1) digits.
Or, we can say that 2n will have (x+1) digits as 2n = 10x.
Therefore, number of digits in 2n = (nlog102) + 1
Below is the implementation of above idea:  

C++




// CPP program to find number of digits
// in 2^n
#include <bits/stdc++.h>
using namespace std;
 
// Function to find number of digits
// in 2^n
int countDigits(int n)
{
    return (n * log10(2) + 1);
}
 
// Driver code
int main()
{
    int n = 5;
    cout << countDigits(n) << endl;
    return 0;
}

Java




// Java program to find number
// of digits in 2^n
import java.util.*;
 
class Gfg
{
    // Function to find number of digits
    // in 2^n
    static int countDigits(int n)
    {
        return (int)(n * Math.log10(2) + 1);
    }
     
    // Driver Code
    public static void main(String args[])
    {
        int n = 5;
        System.out.println(countDigits(n));
    }
}
 
// This code is contributed by Niraj_Pandey.

Python3




# Python3 program to find
# number of digits in 2^n
import math
 
# Function to find number
# of digits in 2^n
def countDigits(n):
    return int(n * math.log10(2) + 1);
 
# Driver code
n = 5;
print(countDigits(n));
 
# This code is contributed
# by mits

C#




// C# program to find number
// of digits in 2^n
using System;
 
class GFG
{
    // Function to find
    // number of digits in 2^n
    static int countDigits(int n)
    {
        return (int)(n * Math.Log10(2) + 1);
    }
     
    // Driver code
    static void Main()
    {
        int n = 5;
        Console.Write(countDigits(n));
    }
}
// This code is contributed by
// Manish Shaw(manishshaw1)

PHP




<?php
// PHP program to find
// number of digits in 2^n
 
// Function to find number
// of digits in 2^n
function countDigits($n)
{
    return intval($n * log10(2) + 1);
}
 
// Driver code
$n = 5;
echo (countDigits($n));
 
// This code is contributed by
// Manish Shaw(manishshaw1)
?>

Javascript




<script>
 
// JavaScript program to find number
// of digits in 2^n
 
    // Function to find number of digits
    // in 2^n
    function countDigits(n)
    {
        return (n * Math.log10(2) + 1);
    }
 
// Driver code
 
        let n = 5;
        document.write(Math.floor(countDigits(n)));
              
             // This code is contributed by souravghosh0416.
</script>

Output: 

2

Time Complexity: O(n)

Auxiliary Space: O(1)




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