Recursive Digit Sum

Super Digit Implementation in Go

Problem statement

We define super digit of a integer x using the following rules:

Given an integer, we need to find the super digit of the integer.

• if x has only one digit, then its super digit is x
• Otherwise the super digit of x is equal to the super digit of the sum of the digits of x.

For example, the super digit of 9875 will be calculated as:

super_digit(9875)    9 + 8 + 7 + 5 = 29 
super_digit(29)      2 + 9 = 11
super_digit(11)      1 + 1 = 2
super_digit(2)       = 2  

Before find the super digit of p we need to calculate p as a product of the given inputs n and k. Let’s illustrate this as follows:

n = '9875'
k = 4

The numer p is created by concatenating the n number k times. 
Therefore p for this concreted example sill be:
p = '9875987598759875'

After calculating the p number we proceed to find their super digit.

 superDigit(p) =  superDigit(9875987598759875)
                   9+8+7+5+9+8+7+5+9+8+7+5+9+8+7+5 = 116
 superDigit(p) =  superDigit(116)
                   1+1+6 = 8
 superDigit(p) =  superDigit(8)

Considerations

We are asked to find the super digit of a given number. The super digit can be calculated by repeatedly adding all the numbers that compounds the p number until there is just one digit.

For starters, instead of concatenating the n number ktimes. Let’s calculate the sum(n) and then multiply it by k.

The super digit, is an algorithm that calculates the digit sum of a number until it becomes a single digit.

The Algorithm

Super digit of p

Resources

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