1. For a natural number $n$ which is not a power of two, denote by $a(n)$ and $b(n)$ the largest and smallest odd divisors of $n$ greater than one, respectively. Determine all numbers $n$ for which

$$
3 a(n) + 5 b(n) = n
$$
1. Let $p$ be the smallest prime divisor of the number $n$ and $n=2^{k} m p$, where $m$ is an odd number and $t \geqslant 0$. Then $a(n)=m p, b(n)=p$ and $2^{k} m p=3 m p+5 p$, from which we get $\left(2^{k}-3\right) m=5$. We have two possibilities.

(i) $m=1$ and $2^{k}-3=5$, i.e., $k=3$. Then $n=8 p$, which is a solution.

(ii) $m=5$ and $2^{k}-3=1$, i.e., $k=2$. Then $n=20 p$, and since $p$ is the smallest odd prime divisor, it must be $p=3$ or $p=5$, which gives two more solutions: $n=60$ and $n=100$.