Number obtained by reducing sum of digits of 2N into a single digit

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Given a positive integer N, the task is to find the single digit obtained after recursively adding the digits of 2^N until a single digit remains.

Examples:

Input: N = 6
Output: 1
Explanation:
2⁶ = 64. Sum of digits = 10.
Now, Sum of digits = 10. Therefore, sum is 1.

Input: N = 10
Output: 7
Explanation: 2¹⁰ = 1024. Sum of digits = 7.

Naive Approach: The simplest approach to solve the problem is to calculate the value of 2^N and then, keep calculating the sum of digits of number until the sum reduces to a single digit.

Time Complexity: O(log(2^N))
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized based on the following observations:

After performing the operation for different values of N, it can be observed that the value repeats after every 6 numbers in the following manner:

If N % 6 = 0, then the single digit sum will be equal to 1.
If N % 6 = 1, then the single digit sum will be equal to 2.
If N % 6 = 2, then the single digit sum will be equal to 4.
If N % 6 = 3, then the single digit sum will be equal to 8.
If N % 6 = 4, then the single digit sum will be equal to 7.
If N % 6 = 5, then the single digit sum will be equal to 5.

Follow the steps below to solve the problem:

If N % 6 is 0 then print 1.
Otherwise, if N % 6 is 1 then print 2.
Otherwise, if N % 6 is 2 then print 7.
Otherwise, if N % 6 is 3 then print 8.
Otherwise, if N % 6 is 4 then print 7.
Otherwise, if N % 6 is 5 then print 5.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the number obtained
// by reducing sum of digits of 2 ^ N
// into a single digit
int findNumber(int N)
{
    // Stores answers for
    // different values of N
    int ans[6] = { 1, 2, 4, 8, 7, 5 };
 
    return ans[N % 6];
}
 
// Driver Code
int main()
{
    int N = 6;
    cout << findNumber(N) << endl;
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
  
// Function to find the number obtained
// by reducing sum of digits of 2 ^ N
// into a single digit
static int findNumber(int N)
{
     
    // Stores answers for
    // different values of N
    int []ans = {1, 2, 4, 8, 7, 5};
 
    return ans[N % 6];
}
 
// Driver Code
public static void main(String args[])
{
    int N = 6;
     
    System.out.println(findNumber(N));
}
}
 
// This code is contributed by ipg2016107

Python3

# Python3 program for the above approach
 
# Function to find the number obtained
# by reducing sum of digits of 2 ^ N
# into a single digit
def findNumber(N):
 
    # Stores answers for
    # different values of N
    ans = [ 1, 2, 4, 8, 7, 5 ]
 
    return ans[N % 6]
 
# Driver Code
if __name__ == "__main__":
  
    N = 6
     
    print (findNumber(N))
 
# This code is contributed by ukasp

C#

// C# program for the above approach
using System;
 
class GFG{
  
// Function to find the number obtained
// by reducing sum of digits of 2 ^ N
// into a single digit
static int findNumber(int N)
{
     
    // Stores answers for
    // different values of N
    int []ans = {1, 2, 4, 8, 7, 5};
 
    return ans[N % 6];
}
 
// Driver Code
public static void Main()
{
    int N = 6;
     
    Console.WriteLine(findNumber(N));
}
}
 
// This code is contributed by mohit kumar 29

Javascript

<script>
// JavaScript program for the above approach
 
// Function to find the number obtained
// by reducing sum of digits of 2 ^ N
// into a single digit
function findNumber(N)
{
 
    // Stores answers for
    // different values of N
    let ans = [ 1, 2, 4, 8, 7, 5 ];
    return ans[N % 6];
}
 
// Driver Code
    let N = 6;
    document.write(findNumber(N) + "<br>");
 
// This code is contributed by Surbhi Tyagi.
</script>

Output:

Time Complexity: O(1)
Auxiliary Space: O(1)

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