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

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==Solution 2==
 
==Solution 2==
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>.
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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>.
  
 
~mathboy100
 
~mathboy100

Revision as of 12:00, 20 February 2023

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?

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

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 $X$ and we can call going $1$ left $Y$. We can build a equation of $5X-3Y=2023$. 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 $2023 = 3\pmod{5}$, 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, so therefore, we must minimize both, which occurs when we jump $4$ to the left and $407$ to the right. The answer is $\boxed{\textbf{(D)}\ 411}$.

~mathboy100

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/1bA7fD7Lg54?t=1723

See Also

2023 AMC 8 (ProblemsAnswer KeyResources)
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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