5. Given a natural number $x=6^{n}+1$, where $n-$ is an odd natural number. It is known that $x$ has exactly three distinct prime divisors, one of which is 11. Find $x$.
Answer: 7777.

Solution. The divisibility of $x$ by 11 is equivalent to $n$ being an odd multiple of 5, that is, $x=6^{5m}+1$ for some odd $m$. From here, using the formula for the sum of a geometric progression,

$$
x=p \cdot q, \quad \text{where } p=6^{m}+1, q=1-6^{m}+6^{2m}-6^{3m}+6^{4m}
$$

Note that $p$ and $q$ are coprime. Indeed, let the number $r$ divide both $p$ and $q$. Then

$$
5=q-(6^{m}+1)-(6^{2m}-1)-(6^{3m}+1)-(6^{4m}-1)
$$

The expressions in parentheses on the right-hand side are divisible by $p$ and, therefore, by $r$. Thus, $r$ is a divisor of 5. But the remainder of the division of $p$ by 5 is 2 for any $m$. Therefore, $p$ is not divisible by 5, which means $r=1$.

Since the number $m$ is odd, one of the prime divisors of $p$ is 7. We will prove that $p$ is a power of 7 only when $m=1$. Suppose $m \geqslant 3$ and $6^{m}+1=7^{s}$. Then the number $7^{s}-1$ is divisible by $2^{m}$ and, therefore, by 8. This implies that $s$ is even, that is, $s=2k$. Therefore, $2^{m} \cdot 3^{m}=(7^{k}-1)(7^{k}+1)$. Since $7^{k}+1$ is not divisible by 3, the number $7^{k}-1$ is divisible by $3^{m}$ and, therefore, $7^{k}-1 \geqslant 3^{m}$. But then

$$
7^{k}+1 \leqslant 2^{m} \cdot 1
$$

1) $p$ is divisible by 11. If $p$ is not a power of 7, we can apply the same reasoning to $p$ as to $x$. In both cases, $p$ is the product of 7 and two coprime factors that are not powers of 7. Therefore, $x$ has at least four different prime divisors, which is impossible.
2) $q$ is divisible by 11. Note that $p$ is divisible by 7 and is not a power of 7, and $q$ is coprime with $p$. Then $q$ must be a power of 11. The remainder of the division of $6^{m}$ by 7 is -1 due to the oddness of $m$. Therefore, the number $q$ gives a remainder of 5 when divided by 8. But the remainders of the division of $11^{s}$ by 7 can only be 4, 2, and 1, so $q$ cannot be a power of 11.