1. Find all positive integers that are not powers of 2 and which are equal to the sum of three times their largest odd divisor and five times their smallest odd divisor greater than 1.
1. Let's use the usual notation $n=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{k}^{\alpha_{k}}$ for the prime factorization of the number $n$, where $p_{1} < p_{2} < \ldots < p_{k}$ are prime numbers, and $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{k}$ are positive integers. The condition $2 \mid n$ implies that $p_{1} = 2$ (since $p_{1} < 2$ is impossible) and that $k \geq 2$ (otherwise, $n$ would be a power of 2). The largest odd divisor of $n$ is clearly the number $p_{2}^{\alpha_{2}} \ldots p_{k}^{\alpha_{k}}$, and its smallest odd divisor (greater than 1) is certainly the prime number $p_{2}$. The equation expressing the condition of the problem can thus be written as

$$
2^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{k}^{\alpha_{k}} = 3 p_{2}^{\alpha_{2}} \ldots p_{k}^{\alpha_{k}} + 5 p_{2} \quad \text{or} \quad \left(2^{\alpha_{1}} - 3\right) p_{2}^{\alpha_{2}-1} \ldots p_{k}^{\alpha_{k}} = 5
$$

(In the case $k=2$, the left side of the modified equation is $\left(2^{\alpha_{1}} - 3\right) p_{2}^{\alpha_{2}-1}$.) Since the number 5 has only two divisors, we have $2^{\alpha_{1}} - 3 \in \{1, 5\}$, so $\alpha_{1} = 2$ or $\alpha_{1} = 3$.

(i) Case $\alpha_{1} = 2$. The modified equation becomes

$$
p_{2}^{\alpha_{2}-1} \ldots p_{k}^{\alpha_{k}} = 5
$$

which is satisfied if and only if either $k=2, p_{2}=5$ and $\alpha_{2}-1=1$, or $k=3$, $\alpha_{2}-1=0$, $p_{3}=5$ and $\alpha_{3}=1$ - in the latter case, however, the prime number $p_{2}$ cannot be arbitrary, because from $22$ there is no such prime number). This case thus corresponds to infinitely many solutions $n = 2^3 p_2^1 = 8 p_2$.

Answer. All positive integers $n$ that satisfy the condition are: $n=60, n=100$ and $n=8 p$, where $p$ is any odd prime number.

Note. The entire process can be written more concisely if we use the expression $n=2^{\alpha} p l$ instead of the "complete" prime factorization of the number $n$, where $2^{\alpha}$ is the highest power of 2 that divides $n$, $p$ is its smallest odd prime factor, and $l$ is an (odd) number that has no prime factor smaller than $p$ (it can be $l=1$, otherwise $l \geq p$). Then our task is to solve the equation

$$
n = 2^{\alpha} p l = 3 p l + 5 p \quad \text{or} \quad \left(2^{\alpha} - 3\right) l = 5.
$$

From this, we have either $l=1$ and $2^{\alpha} - 3 = 5$, or $l=5$ and $2^{\alpha} - 3 = 1$. In the first case, $\alpha=3$, and thus any $n=8 p$ where $p$ is any odd prime number satisfies the condition; in the second case, $l=5$ and $\alpha=2$, so $n=20 p$, but from $5 = l \geq p$ it follows that $p \in \{3, 5\}$, and thus only the numbers $n=60$ and $n=100$ satisfy the condition.

Alternative Solution. From the problem statement, it follows that between the number $n$ and its largest odd divisor $L$, the inequalities $n > 3L$ and $n \leq 3L + 5L = 8L$ hold. Since the ratio $n : L$ is a power of 2, from the inequalities $3L < n \leq 8L$, 2 points for the inequality $n > 3L$, 1 point for the inequality $n \leq 8L$, 1 point for the conclusion about the possible forms $n=4L, n=8L$, and 1 point for their analysis.