2023 AMC 8 Problems/Problem 18
Contents
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 1
We have directions going right or left. We can assign a variable to each of these directions. We can call going right direction and we can call going left . 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 right direction then subtract off in the left 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
Solution 2
Notice that , and jumping to the left increases the value of Greta's position by . Therefore, the number of jumps to the left must be . 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 to the left and to the right. The answer is .
~mathboy100
Animated Video Solution
~Star League (https://starleague.us)
Video Solution by OmegaLearn (Restrictive Counting)
Video Solution by Magic Square
https://youtu.be/-N46BeEKaCQ?t=3673
Video Solution by Interstigation
https://youtu.be/1bA7fD7Lg54?t=1723
Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
Video Solution by harungurcan
https://www.youtube.com/watch?v=Ki4tPSGAapU&t=0s
~harungurcan
See Also
2023 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.