This is a classic Chinese Remainder Theorem. We know that
$$\begin{align*}
n &\equiv 2\pmod{5}\\
n &\equiv 1\pmod{6}.
\end{align*}$$
By the Chinese Remainder Theorem, there is a unique number $n$ modulo $30=5\times 6$ that satisfies both equations.
We can compute it directly: $n=5q+2$ since it leaves a remainder of $2$ when divided by $5$. That means that we must have $5q+2\equiv1\pmod{6}$, which means $5q\equiv -1\equiv 5\pmod{6}$, hence $q\equiv 1\pmod{6}$. So $n=5q+2 = 5(6k+1)+2 = 30k+5+2 = 30k+7$. That is, the solution is $n\equiv 7\pmod{30}$.
Since the number must be between $20$ and $50$, the number is $37$. When divided among $7$ children, she will have $2$ candles left over.
If you must avoid congruences (really, you are only avoiding them explicitly, shoving them "under the carpet"), $n$ must be of the form $6k+1$. We can rewrite this as $n=6k+1 = 5k + (k+1)$. Since, when $n$ is divided by $5$ the remainder is $2$, that means that when we divide $k+1$ by $5$ the remainder is $2$. Therefore, the remainder when dividing $k$ by $5$ must be $1$ (so that the remainder of $k+1$ will be $1+1=2$), so $k=5r+1$. So $n=6k+1 = 6(5r+1)+1 = 30r+7$, and we are back in what we had above.
Let $q_1,q_2,\dots,q_n$ be odd primes of the form $3k+2$. Consider the number $N$, where
$$N=3q_1q_2\dots q_n+2.$$
It is clear that none of the $q_i$ divides $N$, and that $3$ does not divide $N$.
Since $N$ is odd and greater than $1$, it is a product of one or more odd primes. We will show that at least one of these primes is of the form $3k+2$.
The prime divisors of $N$ cannot be all of the shape $3k+1$. For the product of any number of (not necessarily distinct) primes of the form $3k+1$ is itself of the form $3k+1$. But $N$ is not of the form $3k+1$. So some prime $p$ of the form $3k+2$ divides $N$. We already saw that $p$ cannot be one of $q_,\dots,q_n$. It follows that given any collection $\{q_1,\dots,q_n\}$ of primes of the form $3k+2$, there is a prime $p$ of the same form which is not in the collection. Thus the number of primes of the form $3k+2$ cannot be finite.
Best Answer
Without being clever, let's brute force our way through.
We want to understand $2^{2014} - 1 \pmod{1000}$. Notice that $\gcd(3,1000) = 1$, so $3$ has a modular inverse (a quick check shows that it's $667$).
Now, how do we find $2^{2014} - 1 \pmod {1000}$? As $\gcd(2,1000) = 2$, we should worry about the $2$ factor. So we think of $1000$ as $2^3 \cdot 5^3$.
Clearly $2^{2014} \equiv 0 \pmod {2^3}$. And as $\varphi(5^3) = 100$, we know that $2^{2014} \equiv 2^{14} \equiv 9 \pmod {125}$. Putting these together, perhaps with the Chinese Remainder Theorem or by quick brute force (there are only 8 things to check, afterall), we see that $2^{2014} \equiv 384 \pmod{1000}$.
Thus $2^{2014} - 1 \equiv 383 \pmod{1000} $. And thus $\frac{2^{2014}-1}{3} \equiv 667\cdot(2^{2014}-1) \equiv 667 \cdot 383 \equiv 461 \pmod{1000}$.
So the remainder is $461$.