Does positive integer x have a factor f such that 1 < f < x ?

The stem can be rephrased as: Is x prime? With this information we can start looking at the statements:

Statement 1: This is obviously not enough, as there is an infinite number of primes greater than 19!. E.G. 19!+1. As consecutive integers are mutually co-prime, 19!+1 must be a prime number.

Statement 2: Sufficient. This statement explicitly excluded 19!+1, so we need to check if any value x is co-prime with 19!, in which case x would be prime, since 19! contains all factors<=19.

Because of the same reason, we do not need to actively check all values x <=19! + 19, for which x cannot be prime. One example for this: x=19! + 19:
\(x=19! + 19=18!*19+19=(18!+1)*19\) Therefore, x=19!+19 cannot be prime. The same goes for all other x<=19!+19

Now we need to exclude x=19!+20, x=19!+21 and x=19!+22. Since all of the summands added to 19! are non-prime numbers with factors between 1 and 19, they share factors as 19!. Therefore statement 2 is sufficient to conclude that x cannot be prime.

Answer B.