Multiple Persistence (Logical Programs)

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.

For example (Input → Output):

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)

Solution

fun persistence(num: Int) : Int {
    var numToString = num.toString().length
    var numbocc = 0
    if(numToString > 1) {
        numbocc = 1
        var result = num.toString()
        result = passAndRunMutiply(result)
        while(result.length > 1) {
            numbocc++
            result = passAndRunMutiply(result)
        }
    }

    return numbocc
}

fun passAndRunMutiply(num: String): String {
    var result:Int  = 1
    for(i in 0 until num.length) {
        var numfromString = num[i].toString().toInt()
        result = result*(numfromString)
    }
    return result.toString()
}