I guess because adding is not multiplying. (Multiplying is in sense adding, but that's a different story).So if you would multiply by 16 than all the factors would stay the same.
Let me try to reason it out with you. But don't take my word for granted. And excuse me if I do it primitively.
Ok. First of all, when we add number we get a different number with different factors.For example factorise 100 -- 1, 2, 4, 5, 10, 20, 25, 50, 100, now 100+1 - is prime - it has only two factors 1 and 101
To answer your question. So the number X consist of factors \(f_1* f_2 *f_3*f_4... f_n\) when we add another number there are two (I guess) possibilities (I) the number we add may may have common factors with number X, or (II) may not have common factors with number X, (besides 1 of course ) ( I hope I won't go too far in reasoning)
So in first case I guess essentially what is happening is number X with factors \(f_1* f_2 *f_3*f_4... f_n\) is addied with another number with factors lets say \(f_2*f_3\) which is divisible by \(f_2\) (or \(f_3\))because \(\frac{ f_1* f_2 *f_3... f_n+f_2*f_3}{f_2}\) -> \(\frac{f_2(f_1*f_3... f_n+f_3)}{f_2} = f_1*f_3... f_n+f_3\)
(This answers you question what factors - 20! +16 has. Common factors - 1, 2, 4, 8, 16 . So 20 is not a factor while 16 is

))
Hope I didn't mess up. Does that answer your questions?
Should we talk about a second case when there are no common factors?