Consider a set of $$3n$$ numbers having variance 4. In this set, the mean of first $$2n$$ numbers is 6 and the mean of the remaining $$n$$ numbers is 3. A new set is constructed by adding 1 into each of the first $$2n$$ numbers, and subtracting 1 from each of the remaining $$n$$ numbers. If the variance of the new set is $$k$$, then $$9k$$ is equal to ________.

Let the $$3n$$ numbers be $$x_1, x_2, \ldots, x_{3n}$$ with mean of the first $$2n$$ numbers equal to $$6$$ and mean of the remaining $$n$$ numbers equal to $$3$$. The overall mean is $$\bar{x} = \frac{2n \cdot 6 + n \cdot 3}{3n} = \frac{15n}{3n} = 5$$.

The variance is given as $$4$$, so $$\frac{1}{3n}\sum_{i=1}^{3n} x_i^2 - \bar{x}^2 = 4$$, giving $$\frac{1}{3n}\sum x_i^2 = 29$$, hence $$\sum x_i^2 = 87n$$.

Now the new set is formed by adding $$1$$ to each of the first $$2n$$ numbers and subtracting $$1$$ from each of the remaining $$n$$ numbers. The new values are $$y_i = x_i + 1$$ for $$i = 1, \ldots, 2n$$ and $$y_i = x_i - 1$$ for $$i = 2n+1, \ldots, 3n$$.

The new sum is $$\sum y_i = \sum_{i=1}^{2n}(x_i+1) + \sum_{i=2n+1}^{3n}(x_i-1) = \sum x_i + 2n - n = 15n + n = 16n$$. The new mean is $$\bar{y} = \frac{16n}{3n} = \frac{16}{3}$$.

The new sum of squares is $$\sum y_i^2 = \sum_{i=1}^{2n}(x_i+1)^2 + \sum_{i=2n+1}^{3n}(x_i-1)^2 = \sum_{i=1}^{2n}(x_i^2 + 2x_i + 1) + \sum_{i=2n+1}^{3n}(x_i^2 - 2x_i + 1)$$.

This equals $$\sum x_i^2 + 2\sum_{i=1}^{2n}x_i - 2\sum_{i=2n+1}^{3n}x_i + 2n + n = 87n + 2(12n) - 2(3n) + 3n = 87n + 24n - 6n + 3n = 108n$$.

The new variance is $$k = \frac{\sum y_i^2}{3n} - \bar{y}^2 = \frac{108n}{3n} - \frac{256}{9} = 36 - \frac{256}{9} = \frac{324 - 256}{9} = \frac{68}{9}$$.

Therefore $$9k = 9 \cdot \frac{68}{9} = 68$$.