[Solved] this is a program to find prime numbers from 2 to 100 in C [closed]

Your code is looping over all integers from 2 to 99, holding the actual number in i.

for(i = 2; i<100; i++)

Then, for every number i, the code is looping again over all integers from 2 to (i/j).

for(j = 2; j <= (i/j); j++)

Your loop’s finishing condition is mathematically equivalent to j being smaller than the square root of i, since any larger integer j would already contain itself a smaller factor of i.
(To check this, get a paper and rewrite the inequality so hat i is the sole part of the right hand side of your condition.)

Then it checks whether or not j divides i.

(i%j)

If j is a factor of i, then i modulo j is zero and hence

if (!(i%j))

evaluates to true (since 0 is evualted as false and ! negotiates this) and you can break out of the loop since i has a divisor not being 1 or i, hence i is not a prime.

On the other hand, if the inner loop is finished, you have found a prime since it has only 1 and i as divisor.

Needles to say that this bruteforce approach is very slow for large integers (you won’t crack RSA with that), but does this clarify your questions?