Question 499191
Note that 210 = 2*3*5*7. Without loss of generality assume that *[tex \LARGE a_1 \le a_2 \le a_3 \le a_4]. The optimal case would be "a1, a2, a3 are factors of a4" in which the LCM would be simply a4. If this is true, then we can construct casework in which we find the minimum possible value of a4.


If a4 is strictly greater than a3, then a3 is at most 1/2 of a4, and


*[tex \LARGE a_1 + a_2 + a_3 + a_4 \le \frac{1}{2}a_4 + \frac{1}{2}a_4 + \frac{1}{2}a_4 + a_4 = \frac{5}{2}a_4 \Rightarrow \frac{5}{2}a_4 \ge 210 \Rightarrow a_4 \ge 84]. The optimal case would be 84, 42, 42, 42.


The next case would be "a4 = a3" and a2 is strictly less. Then,


*[tex \LARGE a_1 + a_2 + a_3 + a_4 \le \frac{1}{2}a_4 + \frac{1}{2}a_4 + a_4 + a_4 = 3a_4 \Rightarrow 3a_4 \ge 210 \Rightarrow a_4 \ge 70]


Or, if a4 = a3 = a2, then


*[tex \LARGE a_1 + a_2 + a_3 + a_4 \le \frac{1}{2}a_4 + a_4 + a_4 + a_4 = \frac{7}{2}a_4 \Rightarrow \frac{7}{2}a_4 \ge 210 \Rightarrow a_4 \ge 60]


We can check that {30, 60, 60, 60} satisfies, and their LCM is 60.