# 2023 AMC 8 Problems/Problem 18

## Problem

Greta Grasshopper sits on a long line of lily pads in a pond. From any lily pad, Greta can jump 5 pads to the right or 3 pads to the left. What is the fewest number of jumps Greta must make to reach the lilly pad located 2023 pads to the right of her starting position? $\textbf{(A) } 405 \qquad \textbf{(B) } 407 \qquad \textbf{(C) } 409 \qquad \textbf{(D) } 411 \qquad \textbf{(E) } 413$

## Solution 1

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 right direction then subtract off in the left 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 $407 + 4 = \boxed{\textbf{(D)}\ 411}$ as our answer.

~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat

## Solution 2

Notice that $2023 = 3\pmod{5}$, and jumping to the left increases the value of Greta's position $\pmod{5}$ by $2$. Therefore, the number of jumps to the left must be $4 \pmod{5}$. As the number of jumps to the left increases, so does the number of jumps to the right, so therefore, we must minimize both, which occurs when we jump $4$ to the left and $407$ to the right. The answer is $\boxed{\textbf{(D)}\ 411}$.

~mathboy100

## Animated Video Solution

~Star League (https://starleague.us)

## Video Solution (CREATIVE THINKING!!!)

~Education, the Study of Everything

## Video Solution by harungurcan

~harungurcan

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 