2023 SSMO Team Round Problems/Problem 5

Problem

Joshy is playing a game with a dartboard that has two sections. If Joshy hits the first section, he gets $20$ points, and if he hits the second section, he gets $23$ points. Assume Joshy always hits one of the two sections. Let $a$ be the maximum value that Joshy cannot achieve. Let $b$ be the number of positive integer scores Joshy cannot achieve. Let $c$ be the number of ways for Joshy to achieve $2023$ points. Find $(a-b)c$.

Solution

From the Chicken McNugget Theorem, we know: \[a = 20 \cdot 23 - 20 - 23 = 417 \quad \text{and} \quad b = \frac{(20 - 1)(23 - 1)}{2} = 209.\]

Thus, \[a - b = 208.\]

Next, $c$ is the number of ordered integer solutions $(x, y)$ to the equation: \[20x + 23y = 2023.\]

Solving this, we find the general solution: \[x = 23n + 8, \quad y = 81 - 20n.\]

We want integer solutions, so $x \ge 0$ and $y \ge 0$. From $x = 23n + 8 \ge 0$, we get $n \ge -\frac{8}{23} \Rightarrow n \ge 0$. From $y = 81 - 20n \ge 0$, we get $n \le 4$.

So $n = 0, 1, 2, 3, 4$, giving $c = 5$ solutions.

Finally, \[(a - b)c = 208 \cdot 5 = \boxed{1040}.\]

~SMO_Team