5. (10 points) Given that are natural numbers greater than 0, and , if is a multiple of 3, and is a multiple of 5, then the number of different pairs is pairs.
Solution
【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.
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.
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