Each piece has length L/n, and the question is whether L/n < 4, i.e. whether n > L/4.
Statement (1): n is a multiple of 3, but L is unknown, so L/n can be anything. not sufficient.
Statement (2): L = 30 and n ≥ 7, so the piece length is 30/n. At n = 7 the piece is about 4.29 (answer no); at n = 8 it is 3.75 (answer yes). The answer straddles 4. not sufficient.
Together: L = 30 and n is a multiple of 3 greater than 6, so the smallest admissible n is 9. Then the longest possible piece is 30/9 ≈ 3.33 < 4, and every larger n gives an even shorter piece, so the answer is yes for all admissible n. Unique; neither alone.
The smallest-admissible-value check is the lever, adding 'multiple of 3' pushes the smallest n from 7 up to 9, which clears the threshold. Solvers who do not re-enumerate the intersection's extreme may keep the straddle and choose (E), or over-trust (2) and choose (B).