Example 14 (1971 32nd Putnam Mathematical Competition) A person tosses a coin, scoring $a$ points for heads and $b$ points for tails $(a, b$ are positive integers, $a>b)$, and accumulates the score each time. He finds that no matter how many times he tosses, there are 35 scores that cannot be recorded, for example, 58 is one of them. Try to determine the values of $a$ and $b$.
Let a person toss a coin and get heads $x$ times and tails $y$ times, where $x$ and $y$ are non-negative integers.
Thus, the cumulative score is $a x + b y$ points.
First, we prove that $(a, b) = 1$.
Otherwise, if $(a, b) = d > 1$, then $a x + b y$ is always divisible by $d$. This means that all positive integers coprime with $d$ (such as $k d + 1$) are unreachable scores, leading to infinitely many unreachable scores, which contradicts the given condition.
Therefore, $(a, b) = 1$.
Clearly, if $m$ is a reachable score, then the line $a x + b y = m$ must pass through at least one integer point in the first quadrant (including the positive half-axes of the x-axis and y-axis and the origin).
Since $(a, b) = 1$, the numbers
$b \cdot 0, b \cdot 1, \cdots, b \cdot (a-1)$ have distinct remainders when divided by $a$.
This is because if $b i \equiv b j \pmod{a}$, then by $(a, b) = 1$, we have $i - j \equiv 0 \pmod{a}$.
And $|i - j| < a$.
If $b > 1$ and $(a, b) = 1$, then
i.e., $(a, b) = (71, 2), (11, 8), (36, 3)$.
By $(a, b) = 1$, $(36, 3)$ is impossible.
When $a = 71, b = 2$,
$71 \cdot 0 + 2 \cdot 29 = 58$

can be recorded, so $a = 71, b = 2$ is not a valid solution.
Next, consider the integer points on the line $11 x + 8 y = 58$. By
$$
y = \frac{58 - 11 x}{8} = 7 - x + \frac{2 - 3 x}{8}
$$

we can find two adjacent integer points $(6, -1)$ and $(-2, 10)$, which are located in the fourth and second quadrants, respectively. Therefore, 58 cannot be recorded. Thus, $a = 11, b = 8$ is the only solution that meets the requirements of the problem.