Consider the first 100 natural numbers. How many of them are not divisible by any one of 2, 3, 5, 7 and 9?
Answer & explanation
Answer: (c) 22
Divisibility by 9 is already covered by 3, so count the numbers from 1 to 100 that are not divisible by 2, 3, 5 or 7. Below 121 these are just 1 and the primes from 11 to 97, which gives 1 + 21 = 22.
- 9 is a multiple of 3, so only 2, 3, 5 and 7 matter.
- Multiples of 2, 3, 5, 7 up to 100: 50 + 33 + 20 + 14 = 117.
- Multiples of 6, 10, 14, 15, 21, 35: 16 + 10 + 7 + 6 + 4 + 2 = 45.
- Multiples of 30, 42, 70, 105: 3 + 2 + 1 + 0 = 6; multiples of 210: 0.
- Divisible by at least one = 117 − 45 + 6 = 78, so not divisible by any = 100 − 78 = 22.
- Check: such numbers are 1 and the 21 primes from 11 to 97 (the smallest composite without the factors 2, 3, 5, 7 is 11² = 121), so 1 + 21 = 22.
Remember · Drop redundant divisors first (9 with 3). Up to 120, numbers free of 2, 3, 5, 7 are just 1 and primes from 11 onwards.
Question and answer: UPSC's official GS Paper II (2025, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·