Only Few!

Consider the numbers from $100$ to $500$ , which are divisible by 7.

The numbers are $105,112,119,126,........,497.$

Here we use A.P. in order to find the number of terms.

$\large \displaystyle t_n = a + (n-1) d \implies 497 = 105 + (n-1) \times 7.\\ \large \displaystyle \implies n-1 = 56\\ \large \displaystyle \therefore n = 57.$

Therefore, There are $57$ numbers which are divisible by $7$ within $100 - 500.$

Now we come to the numbers divisible by $21.$

The numbers divisible by $21$ are $105,125,........,483.$

Same as the above case. Here also we use A.P.

$\large \displaystyle t_m = a + (m-1)d \implies 483 = 105 + (m-1) \times 21.\\ \large \displaystyle \implies m-1= 19\\ \large \displaystyle \therefore m = 19.$

We require the numbers between $100$ and $500$ which are divisible by $7$ and not by $21$ .

$\large \displaystyle \text{Required solution} = n - m = 57 - 19 = \color{#3D99F6}{\boxed{38}}.$