Problem 5

问题描述：
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?


求出1~20的最小公倍数.

思路如下：

	//辗转相除法求最大公倍数
	public long count_gcd(long a, long b){
		long big =  a>b ? a : b;
		long small = a>b ? b : a;
		long gcd = small;
		long r1, r2;
		while(big%small!=0){
			gcd = big%small;
			big = small;
			small = gcd;
		}
		return gcd;
	}
	
	public long solve1(int number){
		long result = 1;
		long gcd = 1;
		long lcm = 1;
		long a = 2;
		long b;
		for(int i=3; i<=number;i++){
			b = i;
			gcd = count_gcd(a,b);
			lcm = (a*b)/gcd;
			a = lcm;
		}
		return lcm;
	}