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

To solve this, we need to find numbers between 1 and 100 that are not divisible by 2, 3, 5, 7, and 9.

First, notice that any number divisible by 9 is already divisible by 3. Therefore, we only need to find numbers not divisible by the prime numbers 2, 3, 5, and 7.

Numbers not divisible by 2 are odd numbers. There are 50 odd numbers between 1 and 100. From these 50 odd numbers, we must remove those divisible by 3, 5, or 7.

List of odd numbers not divisible by 3, 5, or 7: 1 (1 is not divisible by any of them) Primes greater than 7: 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 (Total 21 numbers) Products of these primes: The only product less than 100 is 11 times 11, which is 121 (too large), so there are no others.

Counting them up: The number 1 plus the 21 prime numbers listed above equals 22.

Therefore, there are 22 such numbers. The correct option is C.