Find f(n) when n=10k, where ff satisfies the recurrence relation f(n)=fn10 with f(1)=10.

Since we need to find $f(n)$ when $n=10k$, we need to determine a pattern when nn is a power of 10.
Substituting $n=10$ into ${\displaystyle f\left(n\right)=f\left(\frac{n}{10}\right)givesf\left(10\right)=f(\frac{10}{10}=f\left(1\right))}$.

Since $f(1)=10$, then $f(10)=10$.
Substituting $n=100$ into $f(n)=f\left(\frac{n}{10}\right)$ gives ${\displaystyle f\left(100\right)=f(\frac{100}{10}=f\left(10\right))}$.

Since $f(10)=10$, then $f(100)=10$.
Substituting $n=1000$ into ${\displaystyle f\left(n\right)=f(\frac{n}{10}givesf\left(1000\right)=f(\frac{1000}{10}=f\left(100\right)))}$.

Since $f(100)=10$, then $f(1000)=10$.
Based on these results, then nn is a power of 10, then the value of f(n) is 10. Therefore, when $n=10k$ we have $f(n)=10.$