How many 2-digit numbers are divisible by 3? What is the sum of all these 2-digit numbers?
Step-by-step solution
Idea: The 2-digit multiples of 3 form an AP 12, 15, …, 99 with d = 3. Find how many terms it has from a + (n − 1)d = 99. For the sum, take out the common factor 3 and use 1 + 2 + … + n = n(n + 1)2.
- Smallest 2-digit multiple of 3 is 12 (= 3 × 4); largest is 99 (= 3 × 33). AP: 12, 15, 18, …, 99 with a = 12, d = 3.½ mark
- 12 + (n − 1) × 3 = 99 ⇒ 3(n − 1) = 87 ⇒ n − 1 = 29 ⇒ n = 30.1 mark
- Sum = 12 + 15 + … + 99 = 3 × (4 + 5 + … + 33) = 3 × [(1 + 2 + … + 33) − (1 + 2 + 3)].1 mark
- = 3 × [33 × 342 − 6] = 3 × (561 − 6) = 3 × 555.1 mark
- = 1665.½ mark
Check: Average of first and last × number of terms: 12 + 992 × 30 = 55.5 × 30 = 1665 ✓.
Answer to write in the exam
2-digit multiples of 3: 12, 15, 18, …, 99 (AP, a = 12, d = 3)
12 + (n − 1) × 3 = 99 ⇒ n − 1 = 29 ⇒ n = 30
∴ 30 two-digit numbers are divisible by 3.
Sum = 3 × (4 + 5 + … + 33) = 3 × (S33 − S3)
= 3 × (33 × 342 − 6) = 3 × (561 − 6) = 3 × 555
∴ Sum = 1665
Common mistakes that cost marks
- Starting the AP at 10 or 3. The first 2-digit multiple of 3 is 12.
- Counting (99 − 12) ÷ 3 = 29 terms and forgetting to add 1 for the first term.
- Using 1 + 2 + … + 33 without removing 1 + 2 + 3 (which are 3, 6, 9, one-digit numbers).
How this can come in the exam
How many 2-digit numbers are divisible by 6?
- 14
- 15
- 16
- 17
Show answer
(B) 15
12, 18, …, 96: 12 + 6(n − 1) = 96 ⇒ n = 15.
Find the sum of all 2-digit numbers divisible by 5.
Show answer
10, 15, …, 95: 18 terms (1 mark). Sum = 5 × (2 + 3 + … + 19) = 5 × (19 × 202 − 1) (1 mark) = 5 × 189 = 945 (1 mark).Try one yourself
How many 2-digit numbers are divisible by 4, and what is their sum?
Show answer
12, 16, …, 96: 22 numbers. Sum = 4 × (3 + 4 + … + 24) = 4 × (300 − 3) = 1188.
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