If the variance of 10 natural numbers 1, 1, 1, ..., 1, $$k$$ is less than 10, then the maximum possible value of $$k$$ is ______.

We have 10 natural numbers: nine 1's and one value $$k$$. The mean is $$\mu = \frac{9 + k}{10}$$.

The variance is $$\sigma^2 = \frac{1}{10}\left[9(1 - \mu)^2 + (k - \mu)^2\right]$$. We compute each deviation: $$1 - \mu = 1 - \frac{9 + k}{10} = \frac{1 - k}{10}$$ and $$k - \mu = k - \frac{9 + k}{10} = \frac{9(k - 1)}{10}$$.

Substituting: $$\sigma^2 = \frac{1}{10}\left[9 \cdot \frac{(k-1)^2}{100} + \frac{81(k-1)^2}{100}\right] = \frac{1}{10} \cdot \frac{(k-1)^2}{100}(9 + 81) = \frac{90(k-1)^2}{1000} = \frac{9(k-1)^2}{100}$$.

We need $$\frac{9(k-1)^2}{100} < 10$$, which gives $$(k-1)^2 < \frac{1000}{9} \approx 111.11$$. Since $$k$$ is a natural number, $$k - 1 \leq 10$$, i.e., $$k \leq 11$$.

Verification: for $$k = 11$$, the variance is $$\frac{9 \times 100}{100} = 9 < 10$$. For $$k = 12$$, the variance is $$\frac{9 \times 121}{100} = 10.89 > 10$$.

The maximum possible value of $$k$$ is $$11$$.