Difference between revisions of "2023 AMC 8 Problems/Problem 18"
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==Video Solution by Dr. David== | ==Video Solution by Dr. David== | ||
https://youtu.be/Qo5H3RVDSbY | https://youtu.be/Qo5H3RVDSbY | ||
| + | |||
| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/OpnNQ4zBA-4 | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2023|num-b=17|num-a=19}} | {{AMC8 box|year=2023|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 10:21, 18 November 2024
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3 (Quick with intuition)
- 5 Solution 3.1
- 6 Video Solution (Solve under 60 seconds!!!)
- 7 Video Solution by Math-X (Smart and Simple)
- 8 Video Solution
- 9 Animated Video Solution
- 10 Video Solution by OmegaLearn (Restrictive Counting)
- 11 Video Solution by Magic Square
- 12 Video Solution by Interstigation
- 13 Video Solution by harungurcan
- 14 Video Solution by Dr. David
- 15 Video Solution by WhyMath
- 16 See Also
Problem
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?
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, we must minimize both, which occurs when we jump to the left and 
to the right. The answer is 
.
~TRALALA
Solution 3 (Quick with intuition)
must be divisible by 3. The smallest value of 
that will achieve this is , which lands it at 
. After that, it takes jumps back, making a total of .
~e___
Solution 3.1
Here is Solution 3 but worded differently:
We can to go back 
steps from 
steps we took past . The other way to say this is we go steps from to get to . What is the least value of ? We need 
to end in a or 
. The least value for , which makes 
, is . 
. Therefore, 
.
Video Solution (Solve under 60 seconds!!!)
https://youtu.be/6O5UXi-Jwv4?si=KvvABit-3-ZtX7Qa&t=826
~hsnacademy
Video Solution by Math-X (Smart and Simple)
https://youtu.be/Ku_c1YHnLt0?si=E5Vs_ZXCVQzSH7Pl&t=3827 ~Math-X
Video Solution
~Education, the Study of Everything
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/DBqko2xATxs&t=2295
Video Solution by harungurcan
https://www.youtube.com/watch?v=Ki4tPSGAapU&t=0s
~harungurcan
Video Solution by Dr. David
Video Solution by WhyMath
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. 
