1134. Armstrong Number

The idea of this solution is to just loop through every digit of n then add the n^(lengthOfNumber) to a variable called sum, then we return sum == n.

func isArmstrong(n int) bool {
    lengthOfNumber := len(strconv.Itoa(n))
    sum := 0
    newN := n

    for newN > 0 {
        sum += powerOfK(lengthOfNumber, newN%10)
        newN /= 10
    }
    return sum == n
}

func powerOfK(lengthOfNumber, n int) int {
    res := 1
    for i := 0; i < lengthOfNumber; i++ {
        res *= n
    }
    return res
}