[LeetCode#263]Factorial Trailing Zeroes

This problem is simple, but you may run into a complexty and easy-wrong way. 
The below is a complex solution(wrong) try to use the same idea from "count primes".

A complex and wrong solution:
public class Solution {
    public boolean isUgly(int num) {
        if (num <= 0)
            return true;
            // throw new IllegalArgumentException("The passed in argument is not legal");
        if (num == 1)
            return true;
        boolean[] check_board = new boolean[num+1];
        Arrays.fill(check_board, true);
        for (int i = 2; i <= Math.sqrt(num); i++) {
            if (check_board[i] == true) {
                for (int j = i*2; j <= num; j = j+i) {
                    check_board[j] = false;
                    if (j == num) {
                        if (!(i == 2 || i == 3 || i== 5)))
                            return true;
                    }
                }
            }
        }
        return false;
    }
}

Why we so many uncessary computing and memeory reasource for a sigle number???
(Something must be wrong for the solution)

If a number is a ugly number, if must consist of (2, 3 5) through following way.
num = (2^i) * (3^j) * (5^k) * 1

Why not we strip out prime factor(2, 3, 5) one by one from num, then check if "num == 1"?
How to strip out prime factor from a integer?
Assume: a is the prime factor
while (num % a == 0) {
    num = num / a;
}
Reason: since num could be fully divided by a (num % a == 0), we could still strip a from num. 
This idea is very tricky compared with our past experience in using array. Take care!

public class Solution {
    public boolean isUgly(int num) {
        if (num <= 0)
            return false;
        if (num == 1)
            return true;
        int[] x = {2, 3, 5};
        for (int a : x) {
            while (num % a == 0) {
                num = num / a;
            }
        }
        return num == 1;
    }
}