The sum and product of the mean and variance of a binomial distribution are 82.5 and 1350 respectively. Then the number of trials in the binomial distribution is

For a binomial distribution with parameters $$n$$ (trials) and $$p$$ (probability of success), the mean is $$\mu = np$$ and the variance is $$\sigma^2 = npq$$ where $$q = 1 - p$$.

We are given that $$\mu + \sigma^2 = 82.5$$ and $$\mu \cdot \sigma^2 = 1350$$. That is, $$np + npq = 82.5$$ and $$np \cdot npq = 1350$$.

From the first equation: $$np(1 + q) = 82.5$$, i.e., $$np(2 - p) = 82.5$$. From the second: $$(np)^2 q = 1350$$, i.e., $$(np)^2(1-p) = 1350$$.

Let $$m = np$$. Then $$m(2-p) = 82.5$$ and $$m^2(1-p) = 1350$$. From the first: $$p = 2 - \frac{82.5}{m}$$, so $$1 - p = \frac{82.5}{m} - 1 = \frac{82.5 - m}{m}$$.

Substituting into the second: $$m^2 \cdot \frac{82.5 - m}{m} = 1350$$, giving $$m(82.5 - m) = 1350$$, so $$82.5m - m^2 = 1350$$, i.e., $$m^2 - 82.5m + 1350 = 0$$.

Multiplying by 2: $$2m^2 - 165m + 2700 = 0$$. Using the quadratic formula: $$m = \frac{165 \pm \sqrt{165^2 - 4 \cdot 2 \cdot 2700}}{4} = \frac{165 \pm \sqrt{27225 - 21600}}{4} = \frac{165 \pm \sqrt{5625}}{4} = \frac{165 \pm 75}{4}$$.

So $$m = \frac{240}{4} = 60$$ or $$m = \frac{90}{4} = 22.5$$.

Case 1: $$m = np = 60$$. Then $$p = 2 - \frac{82.5}{60} = 2 - 1.375 = 0.625 = \frac{5}{8}$$. So $$n = \frac{60}{5/8} = 96$$.

Case 2: $$m = np = 22.5$$. Then $$p = 2 - \frac{82.5}{22.5} = 2 - \frac{11}{3} = -\frac{5}{3}$$, which is negative and invalid.

Hence $$n = 96$$.

Hence, the correct answer is $$\boxed{96}$$.