1.
最大公因數gcd可以用輾轉相除法
ex.陳彥廷的code
   int Ea_a(int a, int b){
	   if(a%b==0)  return b;
	   else  return Ea_b(a%b,b);
       }
   int Ea_b(int a, int b){
	   if(b%a==0)  return a;
	   else  return Ea_a(a,b%a);
       }

2.
最小公倍數lcm(x,y,z)
lcm = x*y*z*gcd(x,y,z)/(gcd(x,y)*gcd(x,z)*gcd(y,z))
    = lcm(x,y)*lcm(x,z)*lcm(y,z)*gcd(x,y,z)/(x*y*z)
    = lcm(lcm(x,y),z)

3.
lcm(x,y) = x*y/gcd(x,y);