To find the numbers between 200 and 500 that are multiples of 3 but not multiples of 12, we'll go through the following steps:

1. Identify numbers that are multiples of 3 between 200 and 500:

   • The smallest multiple of 3 in this range is $201$ (since $200 \div 3 \approx 66.67$).
   • The largest multiple of 3 in this range is $498$ (since $500 \div 3 \approx 166.67$).
   • These numbers form an arithmetic sequence with a common difference of 3. We can calculate the total count using the formula for the number of terms in an arithmetic sequence: $n = \frac{{\text{{last term}} - \text{{first term}}}}{d} + 1$ $n = \frac{{498 - 201}}{3} + 1 = 100$ So, there are 100 numbers that are multiples of 3 between 200 and 500.

2. Identify numbers that are multiples of both 3 and 12 (i.e., multiples of 12) in this range:

   • A number that is a multiple of both 3 and 12 is a multiple of 12. The smallest multiple of 12 in this range is $204$.
   • The largest multiple of 12 in this range is $492$.
   • These numbers also form an arithmetic sequence with a common difference of 12. Let's calculate how many there are: $n = \frac{{492 - 204}}{12} + 1 = 25$ So, there are 25 numbers that are multiples of 12 between 200 and 500.

3. Find numbers that are multiples of 3 but not multiples of 12:

   • Subtract the count of multiples of 12 from the count of multiples of 3: $100 - 25 = 75$

Thus, there are 75 numbers between 200 and 500 that are multiples of 3 but not multiples of 12.

Would you like further details or have any questions? Here are some related questions for practice:

1. How many numbers between 100 and 300 are multiples of 5 but not of 10?
2. Find how many numbers between 1 and 1000 are multiples of 4 but not of 8.
3. What is the sum of all multiples of 7 between 200 and 500?
4. How many numbers between 300 and 600 are multiples of both 3 and 5?
5. Find the smallest multiple of 18 between 400 and 600.

Tip: When working with multiples, breaking the problem into smaller, well-defined steps (like finding sequences) often simplifies the solution.