2023 AMC 8 Problem 18
Attempt Problem 18 of the 2023 AMC 8 below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2023 AMC 8 solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
18.
Greta Grasshopper sits on a long line of lily pads in a pond. From any lily pad, Greta can jump pads to the right or pads to the left. What is the fewest number of jumps Greta must make to reach the lily pad located pads to the right of her starting position?
Answer: D
Small hint:
Start with jumps to the right, which reaches
Big hint:
The extra jumps must have net displacement
Video solution:
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Written solution:
Let be the number of right jumps and be the number of left jumps. We need Reducing modulo gives , so .
To minimize the total number of jumps, use the smallest possible , namely . Then , so .
The fewest number of jumps is .
Thus, D is the correct answer.
Problem 18 in other years
- 1985 AMC 8
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