5. Given a natural number $x=2^{n}-32$, where $n-$ is a natural number. It is known that $x$ has exactly three distinct prime divisors, one of which is 3. Find $x$.
Answer: 480 or 2016.

Solution. Let's write $x$ in the form $32 \cdot N$, where $N=2^{n-5}-1$. One prime divisor of $x$ is 2. Therefore, we need to find all $N$ that have exactly two odd prime divisors, one of which is 3. The divisibility of $N$ by 3 means that $n-5$ is even, i.e., $N=2^{2 m}-1$. If $m=1$, then $N=3$, which does not fit our criteria. For $m=2$ and $m=3$, we get $N=3 \cdot 5$ and $N=3^{2} \cdot 7$, respectively, from which $x=480$ and $x=2016$. We will show that there are no solutions for $m>3$. Note that

$$
N=p \cdot q, \quad \text { where } p=2^{m}-1, \quad q=2^{m}+1
$$

The numbers $p$ and $q$ are coprime, as they are odd and $q-p=2$. One of them is divisible by 3. If $p \vdots 3$, then $m$ is even, and we can factorize $p$ into the product of two coprime numbers, different from 1, in the same way as $N$. Then $N$ is the product of three coprime numbers and, thus, has at least three distinct prime divisors, which is impossible.

Now let $q \vdots 3$. Since the number $q$ is coprime with $p$, it cannot have other prime divisors. Therefore, $q=3^{s}$ and $2^{m}+1=3^{s}$ for some natural $s$. Consider two cases.

1) $s$ is even. Then $s=2 k$, where $k>1$ since $m>3$. Therefore,

$$
2^{m}=\left(3^{k}-1\right)\left(3^{k}+1\right)
$$

In the right-hand side of the equation, there is a product of two consecutive even numbers, both greater than 4. But one of them is not divisible by 4 and therefore is not a power of two, which is impossible.
2) $s$ is odd. Then $s=2 k+1$ for some natural $k$, and

$$
2^{m}-2=3^{2 k+1}-3=3\left(3^{k}-1\right)\left(3^{k}+1\right)
$$

The right-hand side of this equation is divisible by 4, as it contains two even factors, while the left-hand side for $m>3$ is not divisible by 4. Thus, this case is also impossible.