Let $$n$$ be a positive integer. Let $$A = \sum_{k=0}^{n} (-1)^k \cdot {^nC_k}\left[\left(\frac{1}{2}\right)^k + \left(\frac{3}{4}\right)^k + \left(\frac{7}{8}\right)^k + \left(\frac{15}{16}\right)^k + \left(\frac{31}{32}\right)^k\right]$$. If $$63A = 1 - \frac{1}{2^{30}}$$, then $$n$$ is equal to ________.

We have $$A = \sum_{k=0}^{n} (-1)^k \binom{n}{k}\left[\left(\frac{1}{2}\right)^k + \left(\frac{3}{4}\right)^k + \left(\frac{7}{8}\right)^k + \left(\frac{15}{16}\right)^k + \left(\frac{31}{32}\right)^k\right]$$.

This can be split into five sums, each of the form $$\sum_{k=0}^{n}(-1)^k\binom{n}{k}r^k = (1-r)^n$$ by the binomial theorem.

So $$A = \left(1-\frac{1}{2}\right)^n + \left(1-\frac{3}{4}\right)^n + \left(1-\frac{7}{8}\right)^n + \left(1-\frac{15}{16}\right)^n + \left(1-\frac{31}{32}\right)^n$$.

$$A = \frac{1}{2^n} + \frac{1}{4^n} + \frac{1}{8^n} + \frac{1}{16^n} + \frac{1}{32^n} = \frac{1}{2^n} + \frac{1}{2^{2n}} + \frac{1}{2^{3n}} + \frac{1}{2^{4n}} + \frac{1}{2^{5n}}$$.

This is a geometric series with first term $$\frac{1}{2^n}$$ and ratio $$\frac{1}{2^n}$$, so $$A = \frac{\frac{1}{2^n}(1 - \frac{1}{2^{5n}})}{1 - \frac{1}{2^n}} = \frac{1 - \frac{1}{2^{5n}}}{2^n - 1}$$.

Now $$63A = 1 - \frac{1}{2^{30}}$$ gives us $$63 \cdot \frac{1 - \frac{1}{2^{5n}}}{2^n - 1} = 1 - \frac{1}{2^{30}}$$.

Trying $$n = 6$$: $$2^n - 1 = 63$$ and $$5n = 30$$. Then $$63 \cdot \frac{1 - \frac{1}{2^{30}}}{63} = 1 - \frac{1}{2^{30}}$$, which is satisfied.

The answer is $$n = 6$$.