2023 AMC 8 Problems/Problem 18

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 See Also

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

2023 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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All AJHSME/AMC 8 Problems and Solutions

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