Difference between revisions of "2023 AMC 8 Problems/Problem 18"
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| − | We have <math>2</math> directions going <math>5</math> right or <math>3</math> left. We can assign a variable to each of these directions. We can call going right <math>1</math> direction <math>X</math> and we can call going <math>1</math> left <math>Y</math>. We can build a equation of <math>5X-3Y=2023</math>. Where we have to limit the number of moves we do. We can do this by making more of our moves the <math>5</math> move turn then the <math>3</math> 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 <math>2023</math>. The least amount of <math>3</math>’s added to <math>2023</math> to make a multiple of <math>5</math> is <math>4</math> as <math>2023 + 4(3) = 2035</math>. So now we have solved the problem as we just go <math>\frac{2035}{5} = 407</math> hops right, and just do 4 more hops left. Yielding <math>407 + 4 = \boxed{\textbf{(D)}\ 411}</math> as our answer. | + | We have <math>2</math> directions going <math>5</math> right or <math>3</math> left. We can assign a variable to each of these directions. We can call going right <math>1</math> direction <math>X</math> and we can call going <math>1</math> left <math>Y</math>. We can build a equation of <math>5X-3Y=2023</math>. Where we have to limit the number of moves we do. We can do this by making more of our moves the <math>5</math> move turn then the <math>3</math> 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 <math>2023</math>. The least amount of <math>3</math>’s added to <math>2023</math> to make a multiple of <math>5</math> is <math>4</math> as <math>2023 + 4(3) = 2035</math>. So now, we have solved the problem as we just go <math>\frac{2035}{5} = 407</math> hops right, and just do 4 more hops left. Yielding <math>407 + 4 = \boxed{\textbf{(D)}\ 411}</math> as our answer. |
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat | ~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat | ||
Revision as of 10:59, 20 February 2023
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
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. 
