4. The number 1200 is written on the board. Petl appended $10 n+2$ fives to it on the right, where $n-$ is a non-negative integer. Vasya thought that this was the base-6 representation of a natural number $x$, and he factored $x$ into prime factors. It turned out that there were exactly two distinct primes among them. For which $n$ is this possible?
Answer: $n=0$.

Solution. We agree to write the numbers in base 6 in parentheses to distinguish them from decimal numbers. Then

$$
\begin{aligned}
& x=(1200 \underbrace{55 \ldots 5}_{10 n+2})=(1201 \underbrace{00 \ldots 0}_{10 n+2})-1=289 \cdot 6^{10 n+2}-1= \\
& =\left(17 \cdot 6 \cdot 6^{5 n}-1\right)\left(17 \cdot 6 \cdot 6^{5 n}+1\right)=\left(102 \cdot 7776^{n}-1\right)\left(102 \cdot 7776^{n}+1\right)
\end{aligned}
$$

If $n=0$, then $x=101 \cdot 103$, which fits our requirement. Let $n \geqslant 1$. Notice that

$$
102 \bmod 101=1, \quad 7776^{n} \bmod 101=(77 \cdot 101-1)^{n} \bmod 101=(-1)^{n}
$$

Let $a=102 \cdot 7776^{n}-1, b=102 \cdot 7776^{n}+1$. These numbers are coprime, as they are odd and differ by 2. Consider two cases.

1) $n$ is even. Then $a$ is divisible by 101. But $a$ and $b$ have no common prime divisors, so $a=101^{p}$ for some natural number $p$. We get

$$
101^{p}-1=102 \cdot 7776^{n}-2=2\left(51 \cdot 7776^{n}-1\right)
$$

which is impossible, since the left side is divisible by 4, while the right side is not.
2) $n$ is odd. Then $b$ is divisible by 101 and, similarly, $b=101^{q}$ for some natural number $q$. Therefore,

$$
101^{q}-1=102 \cdot 7776^{n}
$$

which is impossible, since the left side is divisible by 5, while the right side is not.