If $$1 + (2 + {}^{49}C_1 + {}^{49}C_2 + \ldots + {}^{49}C_{49})({}^{50}C_2 + {}^{50}C_4 + \ldots + {}^{50}C_{50})$$ is equal to $$2^n \cdot m$$, where $$m$$ is odd, then $$n + m$$ is equal to _____

We need to evaluate $$1 + (2 + {}^{49}C_1 + {}^{49}C_2 + \ldots + {}^{49}C_{49})({}^{50}C_2 + {}^{50}C_4 + \ldots + {}^{50}C_{50})$$.

We know that $${}^{49}C_0 + {}^{49}C_1 + \ldots + {}^{49}C_{49} = 2^{49}$$, so $${}^{49}C_1 + {}^{49}C_2 + \ldots + {}^{49}C_{49} = 2^{49} - 1$$.

Therefore, $$2 + {}^{49}C_1 + \ldots + {}^{49}C_{49} = 2 + 2^{49} - 1 = 2^{49} + 1$$.

Now for the second factor: the sum of even-indexed binomial coefficients of $${}^{50}C_k$$. We know $${}^{50}C_0 + {}^{50}C_2 + {}^{50}C_4 + \ldots + {}^{50}C_{50} = 2^{49}$$. So $${}^{50}C_2 + {}^{50}C_4 + \ldots + {}^{50}C_{50} = 2^{49} - {}^{50}C_0 = 2^{49} - 1$$.

So the expression becomes:

$$1 + (2^{49} + 1)(2^{49} - 1) = 1 + 2^{98} - 1 = 2^{98}$$

So $$2^{98} = 2^n \cdot m$$ where $$m$$ is odd. This means $$n = 98$$ and $$m = 1$$.

Therefore $$n + m = 98 + 1 = 99$$.

Hence, the correct answer is 99.