2023 AMC 8 Problems/Problem 18

Revision as of 03:11, 25 January 2023 by Mayowl (talk | contribs) (Video Solution by OmegaLearn (Using Restrictive Counting))

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

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.


~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, lpieleanu

Animated Video Solution

https://youtu.be/zmRiG52jxpg

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

Video Solution by OmegaLearn (Restrictive Counting)

https://youtu.be/gIjhiw1CUgY

See Also

2023 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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All AJHSME/AMC 8 Problems and Solutions

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