How many three-digit numbers are divisible by 7?
(Hint: All three-digit numbers divisible by 7 form an AP. Find the smallest and largest such three-digit numbers.)
Answer: 105, 112, …, 994 is an AP with d = 7: 105 + (n − 1)7 = 994 gives n = 128.
Step-by-step solution
Idea: List the three-digit multiples of 7 as an AP from the smallest to the largest, then find the position of the largest one.
- Smallest: 100 ÷ 7 = 14 remainder 2, so 7 × 15 = 105. Largest: 999 ÷ 7 = 142 remainder 5, so 7 × 142 = 994.1 mark
- AP: 105, 112, 119, …, 994 with a = 105, d = 7.½ mark
- 105 + (n − 1) × 7 = 994 ⇒ (n − 1) × 7 = 889 ⇒ n − 1 = 127 ⇒ n = 128.1½ marks
128 three-digit numbers are divisible by 7.
Check: Multiples of 7 up to 999: 142; up to 99: 14. 142 − 14 = 128 ✓.
Answer to write in the exam
Smallest 3-digit multiple of 7 = 105; largest = 994
AP: 105, 112, …, 994; a = 105, d = 7
105 + (n − 1) × 7 = 994 ⇒ 7(n − 1) = 889 ⇒ n − 1 = 127
∴ n = 128 three-digit numbers
Common mistakes that cost marks
- Starting at 100 or 98: 100 is not a multiple of 7 and 98 has only two digits.
- Computing (994 − 105) ÷ 7 = 127 and forgetting to add 1 for the first term.
- Taking 999 or 1001 as the largest. 999 is not a multiple of 7 and 1001 has four digits.
How this can come in the exam
MCQ (1 mark)
How many three-digit numbers are divisible by 9?
- 99
- 100
- 101
- 111
Show answer
(B) 100
108 to 999: (999 − 108) ÷ 9 + 1 = 99 + 1 = 100.
Try one yourself
How many three-digit numbers are divisible by 11?
Show answer
110, 121, …, 990: (990 − 110) ÷ 11 + 1 = 81.
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