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

The correct answer is 22.

We are asked to find how many of the first 100 natural numbers (the numbers from 1 to 100, inclusive) are not divisible by any of the numbers 2, 3, 5, 7, or 9.

First, notice that any number divisible by 9 is automatically divisible by 3, since 9 is a multiple of 3. Therefore, if a number is not divisible by 3, it is guaranteed not to be divisible by 9. Thus, checking for non-divisibility by 2, 3, 5, 7, and 9 simplifies to checking for non-divisibility by 2, 3, 5, and 7.

We need to find the count of numbers in the set {1, 2, 3, ..., 100} that are NOT divisible by 2, 3, 5, or 7.

Let us count these numbers directly by listing the prime numbers and 1 in the range 1 to 100:

1. The number 1 is not divisible by any prime, so 1 is included.

2. Any composite number less than or equal to 100 must have a prime factor less than or equal to $\sqrt{100}=10$. The primes less than 10 are 2, 3, 5, and 7.

3. Therefore, any composite number between 1 and 100 is divisible by at least one of 2, 3, 5, or 7.

4. This means the only numbers in the range 1 to 100 that are NOT divisible by 2, 3, 5, or 7 are:
- The number 1
- The prime numbers greater than 7 and less than or equal to 100.

Let us list all the prime numbers between 1 and 100:

Primes less than or equal to 7:
2, 3, 5, 7 (4 primes)

Primes greater than 7 and less than or equal to 100:
11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97

Let us count the primes greater than 7:
There are 21 such prime numbers.

Including the number 1, the total count of natural numbers up to 100 not divisible by 2, 3, 5, 7, or 9 is:
$1+21=22$

Thus, there are 22 natural numbers among the first 100 that are not divisible by 2, 3, 5, 7, or 9.