2023 AMC 8 Problems/Problem 18

Solution

We have $2$ directions going $5$ right or $3$ left. We can assign a variable to each of these directions. We can call going right 1 direction X and we can call going 1 left Y. We can build a equation of $5X-3Y=2023$ . Where we have to limit the number of moves we do. We can do this by making more of our moves the $5$ move turn then the $3$ move turn. The first obvious step is to go some amount of moves in the → direction then subtract off in the ← direction to land on $2023$ . The least amount of $3$ ’s added to $2023$ to make a multiple of $5$ is $4$ as $2023 + 4(3) = 2035$ . So now we have solved the problem as we just go $\frac{2035}{5} = 407$ hops right, and just do 4 more hops left. Yielding $407 + 4 = \boxed{\text{(D)}411}$ as our answer.