Your best bet in a situation like this, where there are relatively few numbers between $50$ and $100$ which are divisible by $7$, is to just list them, and count the number of entries on the list. (Of course, you'd like a more efficient means if the interval over which multiples of a given number span is very very large!) Note: When computing the number of integers between $50$ and $100$ that are divisible by both $7$ and $11$, there is only one such number: $7 \times 11 = 77$.
If, however, the question asked how many such numbers are divisible by $7$ OR $11$, then you'd need to list/count
(1) those divisible by $7$,
(2) those divisible by $11$;
(3) those divisible by both $7$ and $11$: any multiple of $7 \times 11$ in the given range;
Then add the number of entries on list (1) to those on (2), and then subtract the number of entries on (3) from that sum. (Otherwise, without subtracting, you'd be counting, e.g. 77, twice, since it would appear on both (1) and (2).)
In general, if the lower bound of the interval under consideration is $L$, and the upper bound of the interval under consideration is $U$, then the smallest multiple of a positive integer $n$ that is larger than or equal to $L$ is $L/n$, and if the result is not an integer, round up to the next integer (say $k_{min}$). On the other hand, the largest multiple of $n$ that is less than or equal to $U$ is $U/n$, but if the result is not an integer, round down (drop off the remainder), say $k_{max}$. Then, count all multiples of $n$ such that $L \leq nk_{min} < nk_i < nk_{max} \leq U$, where $nk_i$ are all the multiples of $n$ between the least and greatest integers in the given interval [$L, U$].
For every $N$ there is a number $X$ such that $N$ divides $X$ and the sum of digits of $X$ equals $N$.
Proof: Write $N = RM$ where $M$ is coprime to $10$ and $R$ contains only the prime factors $2$ and $5$. Then, by Euler's theorem, $10^{\varphi(M)} \equiv 1 \pmod M$. Consider $X' := \sum_{i=1}^{N} 10^{i\varphi(M)}$. It is a sum of $N$ numbers each of which is congruent to $1$ modulo $M$, so $X' \equiv N\cdot 1 \equiv 0 \pmod M$. Furthermore, the decimal representation of $X'$ contains exactly $N$ ones, all other digits are $0$, so the sum of digits of $X'$ is $N$. Multiply $X'$ by a high power of ten to get a multiple of $R$, call the result $X$. Then $X$ is divisible by $M$ and $R$, hence by $N$, and it has the same digit sum as $X'$ which is $N$.
Best Answer
$$50 = 7 \times 7 +1$$ $$100 = 14 \times 7 +2$$ and the answer is $14-7=7$.
This is effectively what you have done, counting $8 \times 7$, $9 \times 7$, $10 \times 7$, $11 \times 7$, $12 \times 7$, $13 \times 7$, and $14 \times 7$. You will need to be more careful if either of the remainders is $0$.