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Q35·CSAT · Prelims 2025

Numbers not divisible by 2,3,5,7,9 below 100

NumericalDivisibility & CountingNumber theoryHard

Question

Consider the first 100 natural numbers. How many of them are not divisible by any one of 2, 3, 5, 7 and 9?

Options

a

20

b

21

c

22

Answer
d

23

Explanation

Since any number divisible by 9 is already divisible by 3, we only need to find numbers \le 100 that are not divisible by 2, 3, 5, or 7. Numbers coprime to all prime numbers up to ⟨MATH⟩√(100)⟨/MATH⟩ (which is 10) are either the number 1 or prime numbers themselves. The prime numbers greater than 7 and less than 100 are: 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. This gives us exactly 21 prime numbers. Adding the number 1 gives a total count of 21 + 1 = 22.

To find numbers up to N not divisible by any prime \le √(N), count the prime numbers between √(N) and N, and remember to add 1.

Answer: (c).

Question details

Year

2025

Paper

CSAT

Question

Q35

Section

Numerical Ability

Sub-topic

Divisibility & Counting

Type

Number theory

Difficulty

Hard

Source hint

Number theory

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