Let $$X_1, X_2, \ldots, X_{18}$$ be eighteen observations such that $$\sum_{i=1}^{18}(X_i - \alpha) = 36$$ and $$\sum_{i=1}^{18}(X_i - \beta)^2 = 90$$, where $$\alpha$$ and $$\beta$$ are distinct real numbers. If the standard deviation of these observations is 1, then the value of $$|\alpha - \beta|$$ is ______.

From $$\sum_{i=1}^{18}(X_i - \alpha) = 36$$, we get $$\sum X_i - 18\alpha = 36$$, so the mean is $$\bar{X} = \frac{\sum X_i}{18} = \alpha + 2$$.

Expanding $$\sum_{i=1}^{18}(X_i - \beta)^2 = \sum_{i=1}^{18}[(X_i - \bar{X}) + (\bar{X} - \beta)]^2 = \sum(X_i - \bar{X})^2 + 18(\bar{X} - \beta)^2$$.

Since the standard deviation is 1, the variance is 1, so $$\sum(X_i - \bar{X})^2 = 18 \cdot 1 = 18$$.

Substituting into the given equation: $$90 = 18 + 18(\bar{X} - \beta)^2$$, which gives $$(\bar{X} - \beta)^2 = 4$$, so $$\bar{X} - \beta = \pm 2$$.

Since $$\bar{X} = \alpha + 2$$, we have $$\alpha + 2 - \beta = \pm 2$$, giving $$\alpha - \beta = 0$$ or $$\alpha - \beta = -4$$. Since $$\alpha$$ and $$\beta$$ are distinct, $$|\alpha - \beta| = 4$$.