Solution:


If n = \(p^a.q^b.r^c\) has a prime factorization with different prime numbers p, q, and r, then the number of factors of n is (a + 1)(b + 1)(c + 1).

Since 280 = \(2^3·5^1·7^1\), we have n(280) = (3 + 1)(1 + 1)(1 + 1) = 4·2·2 = 16.

Since 30 =\(2^1·3^1·5^1\), we have n(30) = (1 + 1)(1 + 1)(1 + 1) = 2·2·2 = 8.

Then we have \(\frac{280\cdot n\left(x\right)}{30}\) = \(\frac{16\cdot n\left(x\right)}{8}\) = 2n(x) = 12 or n(x) = 6.

We have two cases of integers with 6 factors. They are \(p^2·q^1\) or \(p^5\) where p and q are different prime numbers since (2 + 1)(1 + 1) = 6 and (5 + 1) = 6.

For x = \(p^2·q^1\), when we have p = 2 and q = 3, we have the smallest number \(2^2·3^1\) = 12.

For x = \(p^5\), when we have p = 2, we have the smallest number \(2^5\) = 32.

The smallest value of x is 12.

Therefore, A is the correct answer.

Answer: A