6. Let the function $f(x)=\sin ^{4} \frac{k x}{10}+\cos ^{4} \frac{k x}{10}$, where $k$ is a positive integer. If for any real number $a$, it holds that $\{f(x) \mid a<x<a+1\}=\{f(x) \mid x \in \mathbf{R}\}$, then the minimum value of $k$ is $\qquad$.
Answer: 16.
Solution: From the given conditions, we have $f(x)=\left(\sin ^{2} \frac{k x}{10}+\cos ^{2} \frac{k x}{10}\right)^{2}-2 \sin ^{2} \frac{k x}{10} \cos ^{2} \frac{k x}{10}$
$$
=1-\frac{1}{2} \sin ^{2} \frac{k x}{5}=\frac{1}{4} \cos \frac{2 k x}{5}+\frac{3}{4},
$$

where $f(x)$ reaches its maximum value if and only if $x=\frac{5 m \pi}{k}(m \in \mathbf{Z})$. According to the conditions, any open interval of length 1, $(a, a+1)$, contains at least one maximum point, thus $\frac{5 \pi}{k}5 \pi$.

Conversely, when $k>5 \pi$, any open interval $(a, a+1)$ contains a complete period of $f(x)$, at which point $\{f(x) \mid a<x<a+1\}=\{f(x) \mid x \in \mathbf{R}\}$ holds.
In summary, the smallest positive integer value of $k$ is $[5 \pi]+1=16$.