5. There are six boxes $B_{1}, B_{2}, B_{3}, B_{4}, B_{5}, B_{6}$, and initially each box contains exactly 1 coin. Each time, one of the following two operations can be chosen to perform on them:
(1) Select a box $B_{j}$ $(1 \leqslant j \leqslant 5)$ that contains at least 1 coin, remove 1 coin from box $B_{j}$, and add 2 coins to box $B_{j+1}$.
(2) Select a box $B_{k}$ $(1 \leqslant k \leqslant 4)$ that contains at least 1 coin, remove 1 coin from box $B_{k}$, and swap the contents of boxes $B_{k+1}$ (which may be empty) and $B_{k+2}$ (which may be empty).

Question: Is it possible to perform a series of the above operations such that boxes $B_{1}, B_{2}, B_{3}, B_{4}, B_{5}$ are empty, and box $B_{6}$ contains exactly $2010^{2010^{2010}}$ coins (Note: $\left.a^{b c}=a^{(b c)}\right)$?
5. The answer is affirmative.

Let $A=2010^{2010^{2010}}$.
If the number of coins in boxes $B_{i}, B_{i+1}, \cdots, B_{i+k}$ are $b_{i}, b_{i+1}, \cdots, b_{i+k}$, and after several operations, they become $b_{i}^{\prime}, b_{i+1}^{\prime}, \cdots, b_{i+k}^{\prime}$, we denote this as
$$
\left(b_{i}, b_{i+1}, \cdots, b_{i+k}\right) \rightarrow\left(b_{i}^{\prime}, b_{i+1}^{\prime}, \cdots, b_{i+k}^{\prime}\right)
$$

Thus, we need to perform several operations to achieve
$$
(1,1,1,1,1,1) \rightarrow(0,0,0,0,0, A) \text {. }
$$

First, we prove two lemmas.
Lemma 1 For each positive integer $a$, we have
$$
(a, 0,0) \rightarrow\left(0,2^{a}, 0\right) \text {. }
$$

Proof We use mathematical induction on the positive integer $k (k \leqslant a)$ to prove: $(a, 0,0) \rightarrow\left(a-k, 2^{k}, 0\right)$.

Since $(a, 0,0) \rightarrow(a-1,2,0)=\left(a-1,2^{1}, 0\right)$, the proposition holds for $k=1$.
Assume the proposition holds for $k<a$. Then
$$
\begin{array}{l}
\left(a-k, 2^{k}, 0\right) \rightarrow\left(a-k, 2^{k}-1,2\right) \rightarrow \cdots \\
\rightarrow\left(a-k, 0,2^{k+1}\right) \rightarrow\left(a-k-1,2^{k+1}, 0\right) .
\end{array}
$$
$$
\begin{array}{l}
\text { Hence }(a, 0,0) \rightarrow\left(a-k, 2^{k}, 0\right) \\
\rightarrow\left(a-k-1,2^{k+1}, 0\right),
\end{array}
$$

which means the proposition also holds for $k+1(\leqslant a)$.
Lemma 2 For each positive integer $a$, we have
$$
(a, 0,0,0) \rightarrow\left(0, P_{a}, 0,0\right),
$$

where $P_{n}=2^{2 \cdot \cdot 2}(n$ 2's), and $n$ is a positive integer.
Proof We use mathematical induction on the positive integer $k (k \leqslant a)$ to prove: $(a, 0,0,0) \rightarrow\left(a-k, P_{k}, 0,0\right)$.
By the operation method (1), we have
$$
(a, 0,0,0) \rightarrow(a-1,2,0,0)=\left(a-1, P_{1}, 0,0\right) \text {. }
$$

Thus, the proposition holds for $k=1$.
Assume the proposition holds for $k<a$. Then
$$
\begin{array}{l}
\left(a-k, P_{k}, 0,0\right) \\
\rightarrow\left(a-k, 0,2^{P_{k}}, 0\right)=\left(a-k, 0, P_{k+1}, 0\right) \\
\rightarrow\left(a-k-1, P_{k+1}, 0,0\right) . \\
\text { Hence }(a, 0,0,0) \rightarrow\left(a-k, P_{k}, 0,0\right) \\
\rightarrow\left(a-k-1, P_{k+1}, 0,0\right),
\end{array}
$$

which means the proposition also holds for $k+1(\leqslant a)$.
Returning to the original problem.
Notice that
$$
\begin{array}{l}
(1,1,1,1,1,1) \rightarrow(1,1,1,1,0,3) \\
\rightarrow(1,1,1,0,3,0) \rightarrow(1,1,0,3,0,0) \\
\rightarrow(1,0,3,0,0,0) \rightarrow(0,3,0,0,0,0) \\
\rightarrow\left(0,0, P_{3}, 0,0,0\right)=(0,0,16,0,0,0) \\
\rightarrow\left(0,0,0, P_{16}, 0,0\right) .
\end{array}
$$

And $A=2010^{2010^{2010}}<\left(2^{11}\right)^{2010^{2010}}$
$$
=2^{11 \times 2010^{2010}}<2^{2010^{2011}}<2^{(211)^{2011}}
$$
$$
=2^{211 \times 2011}<2^{2^{15}}<P_{16},
$$

thus the number of coins in box $B_{4}$ is greater than $A$.
$$
\begin{array}{l}
\text { Also, }\left(0,0,0, P_{16}, 0,0\right) \rightarrow\left(0,0,0, P_{16}-1,0,0\right) \\
\rightarrow\left(0,0,0, P_{16}-2,0,0\right) \rightarrow \cdots \rightarrow\left(0,0,0, \frac{A}{4}, 0,0\right), \\
\text { Hence }\left(0,0,0, \frac{A}{4}, 0,0\right) \rightarrow \cdots \rightarrow\left(0,0,0,0, \frac{A}{2}, 0\right) \\
\rightarrow \cdots \rightarrow(0,0,0,0,0, A) .
\end{array}
$$