M7MBA wrote:The integer K is positive, but less than 400. If 21K is a multiple of 180, how many unique prime factors does K have?

A. 1
B. 2
C. 3
D. 4
E. 5

The OA is C.

Experts, may you help me here? I don't have idea how could I solve this PS question. Thanks in advance.

Since 180 = 18 x 10 = 3^2 x 2^2 x 5 and 21K = 3 x 7 x K, we see that K must be a multiple of 3 x 2^2 x 5.

In other words, K = 3 x 2^2 x 5 x n = 60n for for some positive integer n. We see that K already has 3 unique prime factors, namely, 2, 3 and 5. However, since K is less than 400, we see that n can't be more than 6. Because n is no more than 6, we see that K can't have any more prime factors other than 2, 3 and 5.

Answer: C