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Difference between revisions of "2023 SSMO Team Round Problems/Problem 5"

(Created page with "==Problem== Joshy is playing a game with a dartboard that has two sections. If Joshy hits the first section, he gets <math>20</math> points, and if he hits the second section,...")
 
 
Line 3: Line 3:
  
 
==Solution==
 
==Solution==
 +
From the Chicken McNugget Theorem, we know:
 +
<cmath>a = 20 \cdot 23 - 20 - 23 = 417 \quad \text{and} \quad b = \frac{(20 - 1)(23 - 1)}{2} = 209.</cmath>
 +
 +
Thus,
 +
<cmath>a - b = 208.</cmath>
 +
 +
Next, <math>c</math> is the number of ordered integer solutions <math>(x, y)</math> to the equation:
 +
<cmath>20x + 23y = 2023.</cmath>
 +
 +
Solving this, we find the general solution:
 +
<cmath>x = 23n + 8, \quad y = 81 - 20n.</cmath>
 +
 +
We want integer solutions, so <math>x \ge 0</math> and <math>y \ge 0</math>. 
 +
From <math>x = 23n + 8 \ge 0</math>, we get <math>n \ge -\frac{8}{23} \Rightarrow n \ge 0</math>. 
 +
From <math>y = 81 - 20n \ge 0</math>, we get <math>n \le 4</math>.
 +
 +
So <math>n = 0, 1, 2, 3, 4</math>, giving <math>c = 5</math> solutions.
 +
 +
Finally,
 +
<cmath>(a - b)c = 208 \cdot 5 = \boxed{1040}.</cmath>
 +
 +
~SMO_Team

Latest revision as of 21:18, 9 September 2025

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

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