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If the variance of 10 natural numbers $1, 1, 1, \ldots, 1, k$ is less than 10, then the maximum possible value of $k$ is
$(k-1)^2 < \frac{1000}{9}$, $k - 1 < \frac{10\sqrt{10}}{3} \approx 10.54$, so $k \leq 11$.
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