13.19 A person tosses a coin, scoring $a$ points for heads and $b$ points for tails $(a, b$ are positive integers, $a>b$). The scores are accumulated each time. He finds that no matter how many times he tosses the coin, there are 35 scores that cannot be recorded, for example, 58 is one of them. Determine the values of $a$ and $b$.
(32nd Putnam Mathematical Competition, 1971)
[Solution] Let a person 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 $d$, in which case, 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, y-axis, and the origin).
Since $(a, b) = 1$, the numbers
$$
b \cdot 0, b \cdot 1, \cdots, b \cdot (a-1)
$$

these $a$ numbers have distinct remainders when divided by $a$.
This is because if
$$
b i \equiv b j \quad (\bmod a),
$$

then by $(a, b) = 1$, we get $i - j \equiv 0 \quad (\bmod a)$.
And $|i - j| < a$, so the remainders are distinct.
If $a b - a - b = 70$, and $(a, b) = 1$, then
$$
\left\{\begin{array}{l}
a - 1 = 70 \\
b - 1 = 1
\end{array} \text { or } \left\{\begin{array}{l}
a - 1 = 10 \\
b - 1 = 7
\end{array} \text { or } \left\{\begin{array}{l}
a - 1 = 35 \\
b - 1 = 2
\end{array}\right.\right.\right.
$$

Thus, $(a, b) = (71, 2), (11, 8), (36, 3)$.
Since $(a, b) = 1$, $(36, 3)$ is not possible.
For $a = 71, b = 2$,
$$
71 \cdot 0 + 2 \cdot 29 = 58
$$

can be recorded, so $a = 71, b = 2$ is not a valid solution.
Now 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.