Wribreeminsl

2021-01-08

Find $f\left(n\right)$ when $n=10k$, where ff satisfies the recurrence relation $f\left(n\right)=f\frac{n}{10}$ with $f\left(1\right)=10$.

delilnaT

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

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

Since $f\left(10\right)=10$, then $f\left(100\right)=10$.
Substituting $n=1000$ into .

Since $f\left(100\right)=10$, then $f\left(1000\right)=10$.
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 $f\left(n\right)=10.$

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