It is well known that 777-headed dragons have 9 or 13 heads on each neck. Two dragons are the same if they have the same number of necks with 9 heads. How many different 777-headed dragons are there?
I. solution. Let $x$ be the number of 9-headed necks of the 777-headed dragon, and $y$ be the number of 13-headed necks. Then $9 x+13 y=777$, where $x, y \in \mathbb{N}$.

Clearly, $13 y \leq 777$, from which $y \leq\left[\frac{777}{13}\right]=59$. Therefore, $0 \leq y \leq 59$.

$13 y=777-9 x=3(259-3 x)$. Since $3 \mid 3(259-3 x)$, it follows that $3 \mid 13 y$, and thus $(13,3)=1$ implies $3 \mid y$.

259 is not divisible by 3, so 9 is not a divisor of $3(259-3 x)$, and thus not a divisor of $13 y$ either, so $y$ cannot be. Therefore, $y$ can only be of the form $9 k+3$ or $9 k+6$.

1. case: $y=9 k+3 \leq 59$, that is, $0 \leq k \leq 6$.

$$
9 x+13(9 k+3)=777, \quad 9 x+13 \cdot 9 k=738, \quad x+13 k=82, \quad x=82-13 k
$$

This is a positive number for every $0 \leq k \leq 6$.

Thus, the possible $(x ; y)$ pairs for the 9-headed and 13-headed necks are:

$$
(82 ; 3), \quad(69 ; 12), \quad(56 ; 21),(43 ; 30),(30 ; 39),(17 ; 48) \text { and }(4 ; 57) \text {. }
$$

2. case: $y=9 l+6 \leq 59$, that is, $0 \leq l \leq 5$.

$$
9 x+13(9 l+6)=777, \quad 9 x+13 \cdot 9 l=699
$$

The left side is divisible by 9, but the right side is not, which is a contradiction, so this case does not provide a solution.

Therefore, there are a total of seven different 777-headed dragons.

II. solution. Let the number of 9-headed necks be $a$, and the number of 13-headed necks be $b$.

Then the following relationship holds: $9 a+13 b=777$, from which

$$
a=\frac{777-13 b}{9}=86-b+\frac{3-4 b}{9}
$$

The values of $a$ and $b$ are appropriate if both sides are integers and positive; for now, we only deal with the former requirement. This is precisely when $b$ and $\frac{3-4 b}{9}=x$ are both integers, that is, $4 b+9 x=3$, or $b=\frac{3-9 x}{4}=$ $-2 x+\frac{3-x}{4}$. This is an integer if and only if for any integer $y$, $3-x=4 y$, or $x=3-4 y$. This means that

$$
b=-2(3-4 y)+y=-6+9 y \quad \text { and } \quad a=86-(-6+9 y)+(3-4 y)=95-13 y
$$

It is still necessary that neither $a$ nor $b$ be negative: $a=95-13 y \geq 0$, that is, $y \leq 7$, and $b=-6+9 y \geq 0$, or $y \geq 1$. Thus, $y=1,2,3,4,5,6,7$ is possible, so 7 different value pairs result for $(a ; b)$, which means there are 7 different 777-headed dragons.