Challenge 16 - Odd or Even Factors

👋 Welcome Gophers! In this challenge, you will be tasked with implementing a function that will return either “odd” or “even” depending on whether or not a number has an odd or an even number of factors.

View Solution 
package main

import "fmt"
import "math"

func OddEvenFactors(num int) string {
  factors := CheckFactors(num)
  if len(factors) % 2 == 0 {
    return "even"
  }
  return "odd"
}

func CheckFactors(num int) []int {
  factors := []int {1, num}
  limit := int(math.Ceil(float64(num)/2))
  for i := 2; i <= limit; i++ {
    if num % i == 0 {
      factors = append(factors, i) 
    }
  }
  return factors
}

func main() {
  fmt.Println("Odd or Even Factors")

  numFactors := OddEvenFactors(23)
  fmt.Println(numFactors) // "even"

  numFactors = OddEvenFactors(36)
  fmt.Println(numFactors) // "odd"
}

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