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Maximum of sum and product of digits until number is reduced to a single digit
  • Last Updated : 05 Dec, 2018
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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 sum and product will be the same. So, return that number.
  2. Else,
  3. Return maximum of both.

Below is the implementation of above approach:

C++




// CPP implementation of above approach
#include<bits/stdc++.h>
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)<<endl;
        return 0;
    }
// This code is contributed by mits

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
Output:
8

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