MathRevolution wrote:
n(A) denotes the number of positive divisors of a positive integer A. What is the smallest possible value of x satisfying \(\frac{n(280)·n(x)}{ n(30)}\) = 12?
A. 12
B. 14
C. 16
D. 18
E. 20
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 \(\frac{n(280) * n(x)}{n(30)}\) = \(\frac{16n(x)}{8}\) = 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 _________________