2023 AMC 8 Problems/Problem 18
Contents
[hide]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?
Solution
We have directions going right or 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 . Where we have to limit the number of moves we do. We can do this by making more of our moves the move turn then the 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 . The least amount of ’s added to to make a multiple of is as . So now we have solved the problem as we just go hops right, and just do 4 more hops left. Yielding as our answer.
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, lpieleanu
Animated Video Solution
~Star League (https://starleague.us)
Video Solution by OmegaLearn (Restrictive Counting)
See Also
2023 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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All AJHSME/AMC 8 Problems and Solutions |
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