# Maximum of sum and product of digits until number is reduced to a single digit

Last Updated : 24 Aug, 2022

Given a number N, the task is to print the maximum between the sum and multiplication of the digits of the given number until the number is reduced to a single digit.

Note: Sum and multiplication of digits to be done until the number is reduced to a single digit.

Let’s take an example where N = 19,

19 breaks into 1+9=10 then 10 breaks into 1+0=1. 1 is a single digit sum.
Also, 19 breaks into 1*9 = 9. 9 is a single digit multiplication.
So, output is 9 i.e. maximum of 9 and 1.

```Input: N = 631
Output: 8

Input: 110
Output: 2```

Approach:

1. Check if a number is less than 10 then the sum and product will be the same. So, return that number.
2. Else,
3. Return the maximum of both.

Below is the implementation of above approach:

## C++

 `// CPP implementation of above approach` `#include` `using` `namespace` `std;` `    ``// Function to sum the digits until it` `    ``// becomes a single digit` `    ``long` `repeatedSum(``long` `n)` `    ``{` `        ``if` `(n == 0)` `            ``return` `0;` `        ``return` `(n % 9 == 0) ? 9 : (n % 9);` `    ``}`   `    ``// Function to product the digits until it` `    ``// becomes a single digit` `    ``long` `repeatedProduct(``long` `n)` `    ``{` `        ``long` `prod = 1;`   `        ``// Loop to do sum while` `        ``// sum is not less than` `        ``// or equal to 9` `        ``while` `(n > 0 || prod > 9) {` `            ``if` `(n == 0) {` `                ``n = prod;` `                ``prod = 1;` `            ``}` `            ``prod *= n % 10;` `            ``n /= 10;` `        ``}` `        ``return` `prod;` `    ``}`   `    ``// Function to find the maximum among` `    ``// repeated sum and repeated product` `    ``long` `maxSumProduct(``long` `N)` `    ``{`   `        ``if` `(N < 10)` `            ``return` `N;`   `        ``return` `max(repeatedSum(N), repeatedProduct(N));` `    ``}`   `    ``// Driver code` `    ``int` `main()` `    ``{`   `        ``long` `n = 631;` `        ``cout << maxSumProduct(n)<

## Java

 `// Java implementation of above approach` `import` `java.util.*;` `import` `java.lang.*;` `import` `java.io.*;`   `class` `GFG {`   `    ``// Function to sum the digits until it` `    ``// becomes a single digit` `    ``public` `static` `long` `repeatedSum(``long` `n)` `    ``{` `        ``if` `(n == ``0``)` `            ``return` `0``;` `        ``return` `(n % ``9` `== ``0``) ? ``9` `: (n % ``9``);` `    ``}`   `    ``// Function to product the digits until it` `    ``// becomes a single digit` `    ``public` `static` `long` `repeatedProduct(``long` `n)` `    ``{` `        ``long` `prod = ``1``;`   `        ``// Loop to do sum while` `        ``// sum is not less than` `        ``// or equal to 9` `        ``while` `(n > ``0` `|| prod > ``9``) {` `            ``if` `(n == ``0``) {` `                ``n = prod;` `                ``prod = ``1``;` `            ``}` `            ``prod *= n % ``10``;` `            ``n /= ``10``;` `        ``}` `        ``return` `prod;` `    ``}`   `    ``// Function to find the maximum among` `    ``// repeated sum and repeated product` `    ``public` `static` `long` `maxSumProduct(``long` `N)` `    ``{`   `        ``if` `(N < ``10``)` `            ``return` `N;`   `        ``return` `Math.max(repeatedSum(N), repeatedProduct(N));` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String[] args)` `    ``{`   `        ``long` `n = ``631``;` `        ``System.out.println(maxSumProduct(n));` `    ``}` `}`

## Python3

 `# Python 3 implementation of above approach`   `# Function to sum the digits until ` `# it becomes a single digit` `def` `repeatedSum(n):` `    ``if` `(n ``=``=` `0``):` `        ``return` `0` `    ``return` `9` `if``(n ``%` `9` `=``=` `0``) ``else` `(n ``%` `9``)`   `# Function to product the digits ` `# until it becomes a single digit` `def` `repeatedProduct(n):` `    ``prod ``=` `1`   `    ``# Loop to do sum while` `    ``# sum is not less than` `    ``# or equal to 9` `    ``while` `(n > ``0` `or` `prod > ``9``) :` `        ``if` `(n ``=``=` `0``) :` `            ``n ``=` `prod` `            ``prod ``=` `1` `            `  `        ``prod ``*``=` `n ``%` `10` `        ``n ``/``/``=` `10` `    `  `    ``return` `prod`   `# Function to find the maximum among` `# repeated sum and repeated product` `def` `maxSumProduct(N):`   `    ``if` `(N < ``10``):` `        ``return` `N`   `    ``return` `max``(repeatedSum(N), ` `               ``repeatedProduct(N))`   `# Driver code` `if` `__name__ ``=``=` `"__main__"``:`   `    ``n ``=` `631` `    ``print``(maxSumProduct(n))`   `# This code is contributed ` `# by ChitraNayal`

## C#

 `// C# implementation of ` `// above approach` `using` `System;` `class` `GFG ` `{`   `// Function to sum the digits ` `// until it becomes a single digit` `public` `static` `long` `repeatedSum(``long` `n)` `{` `    ``if` `(n == 0)` `        ``return` `0;` `    ``return` `(n % 9 == 0) ? ` `                      ``9 : (n % 9);` `}`   `// Function to product the digits ` `// until it becomes a single digit` `public` `static` `long` `repeatedProduct(``long` `n)` `{` `    ``long` `prod = 1;`   `    ``// Loop to do sum while` `    ``// sum is not less than` `    ``// or equal to 9` `    ``while` `(n > 0 || prod > 9) ` `    ``{` `        ``if` `(n == 0) ` `        ``{` `            ``n = prod;` `            ``prod = 1;` `        ``}` `        ``prod *= n % 10;` `        ``n /= 10;` `    ``}` `    ``return` `prod;` `}`   `// Function to find the maximum among` `// repeated sum and repeated product` `public` `static` `long` `maxSumProduct(``long` `N)` `{`   `    ``if` `(N < 10)` `        ``return` `N;`   `    ``return` `Math.Max(repeatedSum(N), ` `                    ``repeatedProduct(N));` `}`   `// Driver code` `public` `static` `void` `Main()` `{` `    ``long` `n = 631;` `    ``Console.WriteLine(maxSumProduct(n));` `}` `}`   `// This code is contributed` `// by inder_verma`

## Javascript

 ``

Output:

`8`

Time Complexity: O(log10n)

Auxiliary Space: O(1)