Let $$m, n \in N$$ and $$\gcd(2, n) = 1$$. If $$30\binom{30}{0} + 29\binom{30}{1} + \ldots + 2\binom{30}{28} + 1\binom{30}{29} = n \cdot 2^m$$, then $$n + m$$ is equal to ______. (Here $$\binom{n}{k} = {}^nC_k$$)

We need to evaluate $$S = 30\binom{30}{0} + 29\binom{30}{1} + 28\binom{30}{2} + \cdots + 2\binom{30}{28} + 1\binom{30}{29}$$.

This can be written as $$S = \sum_{r=0}^{29}(30-r)\binom{30}{r} = 30\sum_{r=0}^{29}\binom{30}{r} - \sum_{r=0}^{29}r\binom{30}{r}$$.

We know $$\sum_{r=0}^{30}\binom{30}{r} = 2^{30}$$, so $$\sum_{r=0}^{29}\binom{30}{r} = 2^{30} - 1$$.

For the second sum, using the identity $$r\binom{30}{r} = 30\binom{29}{r-1}$$:

$$\sum_{r=0}^{29}r\binom{30}{r} = 30\sum_{r=1}^{29}\binom{29}{r-1} = 30\sum_{j=0}^{28}\binom{29}{j} = 30(2^{29} - 1)$$

since $$\sum_{j=0}^{29}\binom{29}{j} = 2^{29}$$ and $$\binom{29}{29} = 1$$.

Therefore: $$S = 30(2^{30} - 1) - 30(2^{29} - 1) = 30 \cdot 2^{30} - 30 - 30 \cdot 2^{29} + 30 = 30(2^{30} - 2^{29}) = 30 \cdot 2^{29}$$.

Since $$30 = 2 \times 15$$, we get $$S = 15 \cdot 2^{30}$$.

So $$n \cdot 2^m = 15 \cdot 2^{30}$$. Since $$\gcd(2, n) = 1$$, we need $$n$$ to be odd, giving $$n = 15$$ and $$m = 30$$.

Therefore $$n + m = 15 + 30 = 45$$.