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Difference between revisions of "2023 AMC 8 Problems/Problem 18"

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==Problem==
 
==Problem==
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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?
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Greta Grasshopper sits on a long line of lily pads in a pond. From any lily pad, Greta can jump <math>5</math> pads to the right or <math>3</math> pads to the left. What is the fewest number of jumps Greta must make to reach the lily pad located <math>2023</math> pads to the right of her starting position?
  
 
<math>\textbf{(A) } 405 \qquad \textbf{(B) } 407 \qquad \textbf{(C) } 409 \qquad \textbf{(D) } 411 \qquad \textbf{(E) } 413</math>
 
<math>\textbf{(A) } 405 \qquad \textbf{(B) } 407 \qquad \textbf{(C) } 409 \qquad \textbf{(D) } 411 \qquad \textbf{(E) } 413</math>
  
 
==Solution 1==
 
==Solution 1==
−
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>\text{X}</math> and we can call going <math>1</math> left <math>\text{Y}</math>. We can build a equation of <math>5\text{X}-3\text{Y}=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
  
 
==Solution 2==
 
==Solution 2==
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Notice that <math>2023 = 3\pmod{5}</math>, and jumping to the left increases the value of Greta's position <math>\pmod{5}</math> by <math>2</math>. Therefore, the number of jumps to the left must be <math>4 \pmod{5}</math>. 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 <math>4</math> to the left and <math>407</math> to the right. The answer is <math>\boxed{\textbf{(D)}\ 411}</math>.
+
Notice that <math>2023 \equiv 3\pmod{5}</math>, and jumping to the left increases the value of Greta's position <math>\pmod{5}</math> by <math>2</math>. Therefore, the number of jumps to the left must be <math>4 \pmod{5}</math>. 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 <math>4</math> to the left and <math>407</math> to the right. The answer is <math>\boxed{\textbf{(D)}\ 411}</math>.
  
−
~mathboy100
+
~TRALALA
 +
 
 +
 
 +
==Solution 3==
 +
 
 +
<math>5y - 2023</math> must be divisible by 3. The smallest value of <math>y</math> that will achieve this is <math>407</math>, which lands it at <math>2035</math>. After that, it takes <math>4</math> jumps back, making a total of <math>\boxed{\textbf{(D)}\ 411}</math>.
 +
 
 +
 
 +
~e___
  
 
==Video Solution by Math-X (Smart and Simple)==
 
==Video Solution by Math-X (Smart and Simple)==
 
https://youtu.be/Ku_c1YHnLt0?si=E5Vs_ZXCVQzSH7Pl&t=3827 ~Math-X
 
https://youtu.be/Ku_c1YHnLt0?si=E5Vs_ZXCVQzSH7Pl&t=3827 ~Math-X
  
−
 
+
==Video Solution==
−
==Video Solution (CREATIVE THINKING!!!)==
 
 
https://youtu.be/d640itCB9_Y
 
https://youtu.be/d640itCB9_Y
  

Latest revision as of 02:00, 1 February 2024

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 lily pad located $2023$ pads to the right of her starting position?

$\textbf{(A) } 405 \qquad \textbf{(B) } 407 \qquad \textbf{(C) } 409 \qquad \textbf{(D) } 411 \qquad \textbf{(E) } 413$

Solution 1

We have $2$ directions going $5$ right or $3$ left. We can assign a variable to each of these directions. We can call going right $1$ direction $\text{X}$ and we can call going $1$ left $\text{Y}$. We can build a equation of $5\text{X}-3\text{Y}=2023$, where we have to limit the number of moves we do. We can do this by making more of our moves the $5$ move turn then the $3$ 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 $2023$. The least amount of $3$’s added to $2023$ to make a multiple of $5$ is $4$ as $2023 + 4(3) = 2035$. So now, we have solved the problem as we just go $\frac{2035}{5} = 407$ hops right, and just do 4 more hops left. Yielding $407 + 4 = \boxed{\textbf{(D)}\ 411}$ as our answer.

~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat

Solution 2

Notice that $2023 \equiv 3\pmod{5}$, and jumping to the left increases the value of Greta's position $\pmod{5}$ by $2$. Therefore, the number of jumps to the left must be $4 \pmod{5}$. 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 $4$ to the left and $407$ to the right. The answer is $\boxed{\textbf{(D)}\ 411}$.

~TRALALA


Solution 3

$5y - 2023$ must be divisible by 3. The smallest value of $y$ that will achieve this is $407$, which lands it at $2035$. After that, it takes $4$ jumps back, making a total of $\boxed{\textbf{(D)}\ 411}$.


~e___

Video Solution by Math-X (Smart and Simple)

https://youtu.be/Ku_c1YHnLt0?si=E5Vs_ZXCVQzSH7Pl&t=3827 ~Math-X

Video Solution

https://youtu.be/d640itCB9_Y

~Education, the Study of Everything

Animated Video Solution

https://youtu.be/zmRiG52jxpg

~Star League (https://starleague.us)

Video Solution by OmegaLearn (Restrictive Counting)

https://youtu.be/gIjhiw1CUgY

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

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

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