We need to find how many numbers from 1 to 100 are $not$ divisible by 2, 3, 5, 7, or 9.

Notice that:

• $\sqrt{100} = 10$, so any non-prime $\le 100$ must be divisible by a prime $\le 10$.
• The primes $\le 10$ are 2, 3, 5, and 7.

Hence, any number in 1–100 that is $not$ divisible by 2, 3, 5, or 7 must itself be a $prime\ number$ greater than 7 (since smaller ones divide others).

Additionally, $9 = 3^2$, so divisibility by 9 is already covered under 3.

So, we simply count:

• Prime numbers between 1 and 100 = 25
• Exclude 2, 3, 5, and 7 (since they are divisible by themselves): $25 - 4 = 21$
• Include 1, because 1 is $not$ divisible by any of these numbers.

Therefore, total = $21 + 1 = 22$