🇺🇸 AMC 8 ⇄ switch contest
2023 AMC 8

Problem 18

Problem 18 · 2023 AMC 8 Hard
Number Theory divisibilitycasework

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?

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Answer: D — 411 jumps.
Show hints
Hint 1 of 2
Order doesn't matter — jumping right-then-left lands the same place as left-then-right. So all that matters is how many of each.
Still stuck? Show hint 2 →
Hint 2 of 2
She should overshoot 2023 with right jumps (landing on a multiple of 5), then walk back to 2023 in left jumps of 3. Find the first overshoot whose gap is a multiple of 3.
Show solution
Approach: overshoot to a multiple of 5, then step back by 3s
  1. Order doesn't matter, so do all the right jumps first: they land Greta on a multiple of 5 just past 2023. Then left jumps of 3 carry her back to 2023, so the overshoot gap must be a multiple of 3.
  2. Check the multiples of 5 above 2023: 2025 (gap 2, no), 2030 (gap 7, no), 2035 (gap 12 = 3 × 4, yes!).
  3. Right jumps: 2035 ÷ 5 = 407. Left jumps: 12 ÷ 3 = 4. Total: 407 + 4 = 411 jumps. This transfers: ‘reach a target using two step sizes’ problems collapse once you realize order doesn't matter — only the counts of each step do, so you're really just solving 5R − 3L = 2023 in whole numbers.
Another way — pair each left jump with a right jump (MAA):
  1. Pair every left jump with a right jump: a pair nets 5 − 3 = 2 pads forward, and any leftover right jumps add 5 each. With P pairs and Q spare right jumps, 2P + 5Q = 2023 and the total jumps is 2P + Q.
  2. Spare right jumps move farther per jump, so maximize Q (minimize P). The smallest P making 2023 − 2P a multiple of 5 is P = 4, giving Q = (2023 − 8)/5 = 403.
  3. Total jumps: 2·4 + 403 = 411.
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