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

Finding sum of digits of a number until sum becomes single digit

Last Updated : 15 Feb, 2025

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Given an integer n, we need to repeatedly find the sum of its digits until the result becomes a single-digit number.

Examples:

Input: n = 1234
Output: 1
Explanation:
Step 1: 1 + 2 + 3 + 4 = 10
Step 2: 1 + 0 = 1

Input: n = 5674
Output: 4
Explanation:
Step 1: 5 + 6 + 7 + 4 = 22
Step 2: 2 + 2 = 4

Table of Content

[Naive Approach] By Repetitively Adding Digits

The approach is focused on calculating the digital root of a number, which is the result of summing the digits repeatedly until a single-digit value is obtained. Here’s how it works conceptually:

Sum the digits: Start by adding all the digits of the given number.
Check the result: If the sum is a single-digit number (i.e., less than 10), stop and return it.
Repeat the process: If the sum is still more than a single digit, repeat the process with the sum of digits. This continues until a single-digit sum is reached.

C++

// C++ program to find the digit sum by  // repetitively Adding its digits  #include <iostream> using namespace std;  int singleDigit(int n) {     int sum = 0;      // Repetitively calculate sum until     // it becomes single digit     while (n > 0 || sum > 9) {          // If n becomes 0, reset it to sum          // and start a new iteration.         if (n == 0) {             n = sum;             sum = 0;         }          sum += n % 10;         n /= 10;     }     return sum; }  int main() {     int n = 1234;     cout << singleDigit(n);     return 0; }

C

// C program to find the digit sum by  // repetitively Adding its digits  #include <stdio.h>  int singleDigit(int n) {     int sum = 0;      // Repetitively calculate sum until     // it becomes single digit     while (n > 0 || sum > 9) {          // If n becomes 0, reset it to sum          // and start a new iteration.         if (n == 0) {             n = sum;             sum = 0;         }          sum += n % 10;         n /= 10;     }     return sum; }  int main() {     int n = 1234;     printf("%d", singleDigit(n));     return 0; }

Java

// Java program to find the digit sum by  // repetitively Adding its digits  class GfG {     static int singleDigit(int n) {         int sum = 0;          // Repetitively calculate sum until         // it becomes single digit         while (n > 0 || sum > 9) {              // If n becomes 0, reset it to sum              // and start a new iteration.             if (n == 0) {                 n = sum;                 sum = 0;             }              sum += n % 10;             n /= 10;         }         return sum;     }      public static void main(String[] args) {         int n = 1234;         System.out.println(singleDigit(n));     } }

Python

# Python program to find the digit sum by  # repetitively Adding its digits  def singleDigit(n):     sum = 0      # Repetitively calculate sum until     # it becomes single digit     while n > 0 or sum > 9:          # If n becomes 0, reset it to sum          # and start a new iteration         if n == 0:             n = sum             sum = 0          sum += n % 10         n //= 10     return sum  if __name__ == "__main__":     n = 1234     print(singleDigit(n))

C#

// C# program to find the digit sum by  // repetitively Adding its digits  using System;  class GfG {     static int singleDigit(int n) {         int sum = 0;          // Repetitively calculate sum until         // it becomes single digit         while (n > 0 || sum > 9) {              // If n becomes 0, reset it to sum              // and start a new iteration.             if (n == 0) {                 n = sum;                 sum = 0;             }              sum += n % 10;             n /= 10;         }         return sum;     }      static void Main() {         int n = 1234;         Console.WriteLine(singleDigit(n));     } }

JavaScript

// JavaScript program to find the digit sum by  // repetitively Adding its digits  function singleDigit(n) {     let sum = 0;      // Repetitively calculate sum until     // it becomes single digit     while (n > 0 || sum > 9) {          // If n becomes 0, reset it to sum          // and start a new iteration.         if (n === 0) {             n = sum;             sum = 0;         }          sum += n % 10;         n = Math.floor(n / 10);     }     return sum; }  // Driver Code const n = 1234; console.log(singleDigit(n));

Output

Time Complexity: O(log₁₀n), as we are iterating over the digits of the number.
Auxiliary Space: O(1)

[Expected Approach] Using Mathematical Formula

We know that every number in the decimal system can be expressed as a sum of its digits multiplied by powers of 10. For example, a number represented as abcd can be written as follows:

abcd = a*10^3 + b*10^2 + c*10^1 + d*10^0

We can separate the digits and rewrite this as:
abcd = a + b + c + d + (a*999 + b*99 + c*9)
abcd = a + b + c + d + 9*(a*111 + b*11 + c)

This implies that any number can be expressed as the sum of its digits plus a multiple of 9.
So, if we take modulo with 9 on each side,
abcd % 9 = (a + b + c + d) % 9 + 0

This means that the remainder when abcd is divided by 9 is equal to the remainder where the sum of its digits (a + b + c + d) is divided by 9.

If the sum of the digits itself consists of more than one digit, we can further express this sum as the sum of its digits plus a multiple of 9. Consequently, taking modulo 9 will eliminate the multiple of 9, until the sum of digits become single digit number.

As a result, the sum of the digits of any number, will equal its modulo 9. If the result of the modulo operation is zero, it indicates that the single-digit result is 9.
To know about code implementation Refer, Digital Root (repeated digital sum) of the given large integer

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

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Ayush Khanduri

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