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

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==Solution==
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==Problem==
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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 1 direction X and we can call going 1 left Y. 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 → direction then subtract off in the ← 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{\text{(D)}411}</math> as our answer.
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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?
  
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<math>\textbf{(A) } 405 \qquad \textbf{(B) } 407 \qquad \textbf{(C) } 409 \qquad \textbf{(D) } 411 \qquad \textbf{(E) } 413</math>
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==Solution 1==
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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>\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==
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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>.
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~TRALALA
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==Solution 3 (Quick with intuition)==
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<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>.
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~e___
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==Solution 3.1==
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Here is Solution 3 but worded differently:
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We can to go back <math>3x</math> steps from <math>5y</math> steps we took past <math>2023</math>. The other way to say this is we go <math>3x</math> steps from <math>2023</math> to get to <math>5y</math>. What is the least value of <math>3x</math>? We need <math>2023+3x</math> to end in a <math>5</math> or <math>0</math>. The least value for <math>3x</math>, which makes <math>2023+3x=2035</math>, is <math>4</math>. <math>\frac{2035}{5}=407</math>. Therefore, <math>207+4=</math> <math>\boxed{\textbf{(D)}\ 411}</math>.
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==Video Solution (Solve under 60 seconds!!!)==
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https://youtu.be/6O5UXi-Jwv4?si=KvvABit-3-ZtX7Qa&t=826
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~hsnacademy
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==Video Solution by Math-X (Smart and Simple)==
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https://youtu.be/Ku_c1YHnLt0?si=E5Vs_ZXCVQzSH7Pl&t=3827 ~Math-X
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==Video Solution==
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https://youtu.be/d640itCB9_Y
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 +
~Education, the Study of Everything
  
 
==Animated Video Solution==
 
==Animated Video Solution==
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~Star League (https://starleague.us)
 
~Star League (https://starleague.us)
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==Video Solution by OmegaLearn (Restrictive Counting)==
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https://youtu.be/gIjhiw1CUgY
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==Video Solution by Magic Square==
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https://youtu.be/-N46BeEKaCQ?t=3673
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==Video Solution by Interstigation==
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https://youtu.be/DBqko2xATxs&t=2295
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==Video Solution by harungurcan==
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https://www.youtube.com/watch?v=Ki4tPSGAapU&t=0s
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~harungurcan
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==Video Solution by Dr. David==
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https://youtu.be/Qo5H3RVDSbY
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==See Also==
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{{AMC8 box|year=2023|num-b=17|num-a=19}}
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{{MAA Notice}}

Latest revision as of 23:56, 2 November 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 (Quick with intuition)

$5y - 2023$ 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___

Solution 3.1

Here is Solution 3 but worded differently: We can to go back $3x$ steps from $5y$ steps we took past $2023$. The other way to say this is we go $3x$ steps from $2023$ to get to $5y$. What is the least value of $3x$? We need $2023+3x$ to end in a $5$ or $0$. The least value for $3x$, which makes $2023+3x=2035$, is $4$. $\frac{2035}{5}=407$. Therefore, $207+4=$ $\boxed{\textbf{(D)}\ 411}$.

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

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

Video Solution by Dr. David

https://youtu.be/Qo5H3RVDSbY

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. AMC logo.png

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