# #45 - Persistent Bugger - Codewars Kata (6 kyu)

## Instructions

Write a function, persistence, that takes in a positive parameter num and returns its multiplicative persistence, which is the number of times you must multiply the digits in num until you reach a single digit.

Example

39 --> 3 (because 3*9 = 27, 2*7 = 14, 1*4 = 4 and 4 has only one digit)
999 --> 4 (because 9*9*9 = 729, 7*2*9 = 126, 1*2*6 = 12, and finally 1*2 = 2)
4 --> 0 (because 4 is already a one-digit number)

## My solution:

``````function persistence(num) {
let count = 0;
while(num.toString().length > 1){
num = num.toString().split('').reduce((acc,el)=>acc *= +el,1)
count++
}
return count
}
``````

## Explanation

First I created the variable count, where I'll save the number of times the number had to be multiplied so it gets to only one digit

``````let count = 0;
``````

Then I used a while loop that will iterate until the length of the number is 1

``````  while(num.toString().length > 1)
``````

Inside of that while loop, I revalued the variable "num" to save the result of the multiplication of the digits, I did this by splitting the number into an array, and then I just used the reduce function to save the result of the multiplication of the digits.

Note: In the next iteration of the loop the variable "num" will be the result of the multiplication from the last digits

After that I just added one to the count variable

``````  num = num.toString().split('').reduce((acc,el)=>acc *= +el,1)
count++
``````

At the end I just returned count

``````  return count
``````