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

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 = 6Output:1Explanation:

2^{6}= 64. Sum of digits = 10.

Now, Sum of digits = 10. Therefore, sum is 1.

Input:N = 10Output:7Explanation:2^{10}= 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:**

1

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