5. (10 points) Given that $x, y$ are natural numbers greater than 0, and $x+y=150$, if $x$ is a multiple of 3, and $y$ is a multiple of 5, then the number of different pairs $(x, y)$ is $\qquad$ pairs.
【Analysis】First, according to the divisibility rule of 5, the last digit must be 0 or 5. Therefore, the difference between 150 and a multiple of 5 will still have a last digit of 0 or 5. We can enumerate the numbers that end in 0 or 5.

【Solution】Solution: According to the divisibility rule of 5, the last digit must be 0 or 5. Therefore, 150 minus this number will still have a last digit of 0 or 5. We can find the numbers ending in 0 or 5 that are multiples of 3.
$30, 60, 90, 120, 15, 45, 75, 105, 135$ There are 9 numbers that satisfy the condition.
Thus, there are 9 pairs of numbers.
The answer is: 9.
【Comment】This problem tests the divisibility properties of numbers. The key is to find the numbers ending in 0 or 5 that are multiples of 3, and then enumerate to solve the problem.