Step 1: Understanding the problem.
To find the number of factors of 1080 that contain exactly one trailing zero, we need to consider the prime factorization of 1080 and how trailing zeros are formed. Trailing zeros are created when a number has factors of 10, which in turn can be broken down into factors of 2 and 5.
Step 2: Prime factorization of 1080.
First, find the prime factorization of 1080:
\[ 1080 = 2^3 \times 3^3 \times 5 \]
Step 3: Conditions for exactly one trailing zero.
A factor of 1080 will have exactly one trailing zero if it contains one factor of 2 and one factor of 5. This is because the presence of one 2 and one 5 forms the number 10, which is responsible for the trailing zero. Therefore, the number of factors of 1080 that contain exactly one trailing zero is the number of ways we can choose powers of 2, 3, and 5 that form a factor of the given number.
- The power of 2 can be 1, 2, or 3.
- The power of 3 can be 0, 1, 2, or 3.
- The power of 5 must be 1.
Step 4: Counting the factors.
The total number of factors of 1080 containing exactly one trailing zero is:
\[
3 \times 4 = 12
\]
But since only one power of 5 is allowed, we have \( 3 \times 4 = 12 \) factors that contain exactly one trailing zero.
Step 5: Conclusion.
The correct answer is (B) 6.