Step-by-Step Solution
Key Concept: The three‑digit numbers divisible by 7 form an arithmetic progression (AP) with common difference 7. By finding the first and last terms of this AP and using the formula \(n = \frac{l-a}{d}+1\), we can count the total number of such numbers.
1. Identify the smallest three‑digit multiple of 7\
\(100 \div 7 = 14\) remainder 2, so the next multiple is \(7\times 15 = 105\).\
Hence, the first term \(a = 105\).
2. Identify the largest three‑digit multiple of 7\
\(999 \div 7 = 142\) remainder 5, so the greatest multiple not exceeding 999 is \(7\times 142 = 994\).\
Hence, the last term \(l = 994\).
3. Common difference\
Since we are dealing with consecutive multiples of 7, the common difference \(d = 7\).
4. Number of terms in the AP\
Use the formula for the number of terms of an AP:
$$n = \frac{l - a}{d} + 1$$
Substituting the values:
$$n = \frac{994 - 105}{7} + 1 = \frac{889}{7} + 1 = 127 + 1 = 128$$
5. Conclusion\
Therefore, there are \(128\) three‑digit numbers that are divisible by 7.
Correct Answer: 128