Lord Moneybag said to his grandson: "Bill, pay attention! Christmas is coming soon. I have taken a sum between 300 and 500 pounds, which is a multiple of 6. You will get 5 pounds in 1-pound coins. When I give you one pound at a time, the remaining amount will first be divisible by 5, then by 4, then by 3, then by 2, and finally only by 1 and itself. If you can tell me how much money I have, you'll get an extra ten pounds." How much money did the lord take with him?
I. Solution. Let's assume that the lord took $n$ pounds with him. According to the problem, the following divisibilities hold: $6|n, 5| n-1, 4|n-2, 3| n-3, 2 \mid n-4$, and we also know that $n-5$ is a prime number. From the condition $5 \mid n-1$, it follows that $n-1$ ends in either 0 or 5, so $n$ must end in either 1 or 6. Since $6 \mid n$, it follows that $n$ is even, so it can only end in 6. Considering that the lord took an amount between 300 and 500 pounds, $n$ can only be of the form $\overline{3 a 6}$ or $\overline{4 b 6}$. Since $n$ is divisible by 6, it is also divisible by 3, so the sum of its digits must be divisible by 3. Therefore, the value of $n$ can only be one of the numbers 306, 336, 366, 396, 426, 456, and 486. The condition $4 \mid n-2$ is only satisfied by the numbers 306, 366, 426, and 486.

Forming the possible values of $n-5$:

$$
\begin{aligned}
& 306-5=301=7 \cdot 43 \\
& 366-5=361=19^{2} \\
& 426-5=421 \text { prime number } \\
& 486-5=481=13 \cdot 37
\end{aligned}
$$

Thus, the only possible value of $n$ is 426, so the lord took 426 pounds with him.

II. Solution. Let $x$ be the amount the lord took with him. We know that $6|x, 5| x-1, 4 \mid x-2$, $3|x-3, 2|x-4$, and $x-5$ is a prime number. From these, it follows that $x-6$ is divisible by 6, 5, 4, 3, and 2, so it is certainly divisible by $[6 ; 5 ; 4 ; 3 ; 2]=60$. Four multiples of 60 fall into the appropriate interval: 300, 360, 420, and 480. Among these, we are looking for the one that, when 1 is added, results in a prime number. $301=7 \cdot 43, 361=19^{2}, 481=13 \cdot 37$, only 421 is prime.

Therefore, the lord took $421+5=426$ pounds with him.