2023 AMC 8

Complete problem set with solutions and individual problem pages

Problem 18 Hard

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?

  • A.

    405

  • B.

    407

  • C.

    409

  • D.

    411

  • E.

    413

Answer:D

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 an 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 than 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.

 

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}.