Find the asymptotic bound $\Theta$ of $t(n)=t(\frac{n}{5})+t(\frac{n}{17})+n$
If we are not restricted by "using the master theorem", then either a better version of the master theorem, the versatile Akra-Bazzi method or the elementary way to show many recurrence relations mean $\Theta(n)$ or $\Theta(n\ln n)$ is an easy approach.

Otherwise, let us see how we can use the master theorem. We assume here $t(n)\geq 0$ for all $n \geq 0$, although the final asymptotic answer is actually valid without that non-negativity restriction. Let $c$, $m$ be two positive constants such that $t(n) \lt m$ for all $n &lt; c$.

To bound $t$ from below, let us define function $t_1$ by
  $$t_1(n) = \begin{cases}
    0 &amp; n &lt;= c\\
    t_1\left(\frac{n}{5}\right)  + n &amp; n\gt c
  \end{cases}$$

To bound $t$ from above, let us define function $t_2$ by
  $$t_2(n) = \begin{cases}
    m &amp; n &lt;= c\\
    t_2\left(\frac{n}{5}\right) + (t_2\left(\frac{n}{5}\right) + m) + n &amp; n\gt c
  \end{cases}$$

The rest of this solution will be proving $t_1(n)\leq t(n)\leq t_2(n)$ for all $n\ge 0$ by  induction followed by applying the master theorem to $t_1$ and $t_2$ to show that $t(n) = \Theta(n)$. It is left as an easy exercise for interested readers.