How many 3-digit numbers are divisible by 6 in all?

Concept: The multiples of any number k form an arithmetic progression (AP) with first term k and common difference k. To find how many terms of an AP lie in a range, use the n-th term formula an = a1 + (n − 1) × d, where a1 is the first term inside the range, d is the common difference, and n is the number of terms up to the last term an.

Application: Apply this to the 3-digit multiples of 6.

1. The 3-digit numbers range from 100 to 999.

2. First 3-digit multiple of 6: 100 ÷ 6 = 16.67, so round up to 6 × 17 = 102.

3. Last 3-digit multiple of 6: 999 ÷ 6 = 166.5, so round down to 6 × 166 = 996.

4. Using a1 = 102, d = 6, and last term an = 996 in the n-th term formula: 996 = 102 + (n − 1) × 6, so (n − 1) = 894 ÷ 6 = 149, giving n = 150.

Cross-check: Count all multiples of 6 up to 999 and remove the multiples of 6 that are not 3-digit (up to 99): ⌊999 ÷ 6⌋ − ⌊99 ÷ 6⌋ = 166 − 16 = 150, matching the count found above.

Answer: There are 150 three-digit numbers divisible by 6.