There are 300 three-digit numbers divisible by 3.
Detailed Explanation
The question asks for the number of three-digit numbers that are divisible by 3 and also form an arithmetic progression. A sequence of numbers is considered to be an arithmetic progression if the difference between consecutive terms is constant. Three-digit numbers range from 100 to 999. We need to determine how many of these are divisible by 3.
Determining the number of three-digit multiples of 3:
- The smallest three-digit number divisible by 3 is 102 (3 x 34).
- The largest three-digit number divisible by 3 is 999 (3 x 333).
- All multiples of 3 form an arithmetic progression with a common difference of 3 (e.g., 102, 105, 108, ...).
Calculating the number of terms in the sequence:
As per the provided reference:
- The number of three-digit numbers divisible by 3 is calculated as (999 - 102)/3 + 1.
- (999 - 102)/3 = 897/3 = 299
- 299 + 1 = 300
Therefore, there are 300 three-digit numbers that are divisible by 3.
Important Note: Since every number divisible by 3 is a multiple of 3, any sequence of multiples of 3 will inherently be an arithmetic progression, where the common difference is 3. Therefore, all 300 three-digit numbers that are multiples of 3 are in arithmetic progression.
Conclusion
There are 300 three-digit numbers that are divisible by 3, and they form an arithmetic progression.
