Task 1. For a positive integer $n$ that is not a power of two, we define $t(n)$ as the largest odd divisor of $n$ and $r(n)$ as the smallest positive odd divisor of $n$ that is not equal to 1. Determine all positive integers $n$ that are not powers of two and for which

$$
n=3 t(n)+5 r(n)
$$
Solution I. If $n$ is odd, then $t(n)=n$, so $3 t(n)$ is greater than $n$, contradiction. If $n$ is divisible by 2 but not by 4, then $t(n)=\frac{1}{2} n$ and $3 t(n)$ is again greater than $n$, contradiction again. We can conclude that $n$ must be divisible by 4 at least. If $n$ is divisible by 16, then $t(n) \leq \frac{1}{16} n$. Furthermore, $r(n) \leq t(n)$, so $3 t(n)+5 r(n) \leq 8 t(n) \leq \frac{1}{2} n < n$, contradiction. Therefore, $n$ is not divisible by 16. We can thus write $n=4 m$ or $n=8 m$ with $m \geq 3$ odd.

Suppose $n=4 m$ with $m \geq 3$ odd. Then $t(n)=m$, so $4 m=3 m+5 r(n)$, hence $5 r(n)=m$. Since $r(n)$ is equal to the smallest odd prime divisor of $n$, which is also the smallest prime divisor of $m$, $m$ must be of the form $m=5 p$ with $p \leq 5$ an odd prime number. Thus, $m=15$ or $m=25$, which gives $n=60$ or $n=100$. Both solutions are valid.

Suppose that $n=8 m$ with $m \geq 3$ odd. Again, $t(n)=m$, so $8 m=3 m+5 r(n)$, hence $5 r(n)=5 m$, or $r(n)=m$. We see that $m$ is prime. So $n=8 p$ with $p$ an odd prime number. This family of solutions is also valid.

We find the solutions to be $n=60, n=100$ and $n=8 p$ with $p$ an odd prime number.