Counting multiples (divisibility by 6): How many three-digit numbers (from 100 to 999) are exactly divisible by 6?

Introduction / Context:
A number divisible by 6 must be divisible by both 2 and 3, i.e., it must be a multiple of 6. Counting the multiples of a fixed integer in a closed interval can be done using floor division endpoints.

Given Data / Assumptions:

• Range: 100 to 999 inclusive.
• Divisor: 6.

Concept / Approach:
The count of multiples of k in [A, B] is floor(B/k) − floor((A−1)/k). Apply with A = 100, B = 999, k = 6.

Step-by-Step Solution:

Verification / Alternative check:
First multiple ≥ 100 is 102 (6 * 17); last multiple ≤ 999 is 996 (6 * 166). Number of terms in the AP 102, 108, …, 996 is (166 − 17 + 1) = 150.

Why Other Options Are Wrong:

• 149 or 151: Off by one due to endpoint mishandling.
• 166: That is floor(999/6); ignores the 1–99 block.

Common Pitfalls:
Using B/k − A/k without floors or forgetting to subtract the multiples below A.

Final Answer: