REM = Remainder.
Now , It is easy to think of doing it by dividing the numbers starting from 100 till whatever until you get remainder as 11. But isthere a way to solve it by formula.
Like , what I have doing is and not getting the right answer for it.
$\frac{100 + x}{13}$= y (for y it is Some value but we don’t know)and (100+x because it has to be a number either greater or or equal to 100)
Then , 100+x - 13(y) = 11.
I’m not getting how to solve further than this.
One way I tried is this but got wrong ans.
So , from here.
100 + x - 13(6) = 9 + x.
9+x = 9+x
But x gets cancel out.

Firstly, the correct notation for remainder $REM(n \div p) = r$ is
$$n = pq + r \implies n\equiv r \mod p$$
Examples include $13\equiv 3\mod 5$ and $444 \equiv 3 \mod 7$
So you want to find the minimum $n\geq 100$ such that $n\equiv 11\mod 13$
$$n\equiv 11\mod 13, n\geq 100$$
$$n + 100 \equiv 11 \mod 13, n \geq 0$$
$$n+9+13 \cdot 7 \equiv 11\mod 13,n\geq 0$$
Since adding 13 won't change the remainder, we can ignore the $13\cdot 7$ term:
$$n+9\equiv 11\mod 13, n\geq 0$$
$$n\equiv 2\mod 13$$
In english terms, this means $n$ (the two-digit part) has remainder 2 when divided by $13$. The trivial example is $n=2 \implies 102\equiv 11\mod 13$, but other examples are $n=15,28,41,54,\cdots$. These give $115,128,141,154,\cdots$ which all give remainder of $11$ when divided by $13$.
Hope this helps!