Project Euler Problem 5: Smallest multiple

Problem

 2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.

 What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?

Solution

Attempt 1: Count the occurrences of the prime factor for each number, and multiply them all.

      private static Map<Integer, Integer> primeNumberOccurrence = new HashMap<>();  
      public static void main(String[] args) {  
           for (int i=2; i<=20; i++){  
                updatePrimeNumberOccurrence(i);  
           }  
           int total = 1;  
           for (int key : primeNumberOccurrence.keySet()){  
                int occurrence = primeNumberOccurrence.get(key);  
                total *= Math.pow(key, occurrence);  
           }  
           System.out.println(total);  
      }  
      private static void updatePrimeNumberOccurrence(int number){  
           int divisor = 2;  
           int divisorCount = 0;  
           while(number > 1){  
                if(number % divisor == 0){  
                     number = number / divisor;  
                     divisorCount++;  
                }else{  
                     update(divisor, divisorCount);  
                     divisor++;  
                     divisorCount = 0;  
                }  
           }  
           update(divisor, divisorCount);  
      }  
      private static void update(int divisor, int divisorCount){  
           if (divisorCount != 0){  
                int oldCount = primeNumberOccurrence.get(divisor) != null ? primeNumberOccurrence.get(divisor) : 0;  
                if (divisorCount > oldCount){  
                     primeNumberOccurrence.put(divisor, divisorCount);  
                }  
           }  
      }  

Attempt 2: Use greatest common factor

      private static final int NUMBER = 20;  
      public static void main(String[] args) {  
           int total = NUMBER;  
           for (int i = NUMBER-1; i >= 2; i--){  
                //greatest common factor  
                total *= i / BigInteger.valueOf(total).gcd(BigInteger.valueOf(i)).intValue();  
           }  
           System.out.println(total);  
      }