AMC 8 · 2023 · #18

Grade 5 number-theoryarithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint

linear-diophantinedivisibility-rulesoptimization-counting guess-and-checkmodular-arithmeticbound-inequality-then-enumerate ↑ Prerequisites: divisibility-rulesmulti-digit-arithmetic

📏 Medium solution 💡 3 insights

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Problem

Greta sits on a long row of lily pads. From any pad she can jump exactly +5 pads (right) or -3 pads (left). What is the smallest number of jumps that lands her exactly 2023 pads to the right of her start?

Pick an answer.

(A)

405

(B)

407

(C)

409

(D)

411

(E)

413

AMC 8 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Guess and Check

If Greta could only jump right, she would need 2023 / 5 = 404.6 jumps — not a whole number, so right-only is impossible. The next idea is to try the smallest values of R that put 5R just above 2023 and see how many left jumps fill the gap. Tool #6 (Guess and Check) is perfect: guess R = 405, 406, 407, … and check whether the leftover 5R - 2023 is a multiple of 3. Tool #9 (Easier Problem) tells us to think of this as "how far past 2023 do I overshoot, and is that overshoot built out of 3s?" Tool #5 (Pattern) confirms the divisibility-by-3 rule on the overshoot.

1STEP 1

Each right jump adds 5, each left subtracts 3; to finish at +2023 the moves must balance to 5R - 3L = 2023.

5R - 3L = 2023, R ≥ 0, L ≥ 0

2STEP 2

Left jumps only subtract, so right jumps alone must reach 2023; since 5 × 404 = 2020 is under 2023, we need R ≥ 405.

5 × 404 = 2020 ＜ 2023 ＜ 2025 = 5 × 405 → R ≥ 405

3STEP 3

Try R = 405: 5R = 2025, overshoot 2. But 3L = 2 gives no whole-number L (2 isn't a multiple of 3). Reject.

5(405) - 3L = 2023 → 3L = 2 → no whole-number L

4STEP 4

Try R = 406: 5R = 2030, overshoot 7. 7 isn't a multiple of 3 either. Reject.

5(406) - 3L = 2023 → 3L = 7 → no whole-number L

5STEP 5

Try R = 407: 5R = 2035, overshoot 12 = 3 × 4, so L = 4 works — total R + L = 411.

5(407) - 3L = 2023 → 3L = 12 → L = 4 → R + L = 411

6STEP 6

Every extra 3 right jumps needs 5 more left jumps, so the total climbs by 8 — no later solution beats it. Minimum is 411, choice (D).

R + L = 411 → (D) 411

Answer

411

If Greta only jumped right, she would need

\frac{2023}{5}

= 404.6 jumps — between 404 and 405. Since 5 and 3 don't share factors with 2023 in a clean way, she has to add a few left jumps to make the arithmetic work, and the smallest correction is L = 4 together with R = 407. The answer 411 is just 4 more than the natural floor of 407, which matches our four extra left jumps. All other answer choices (405, 407, 409, 413) either ignore the left jumps entirely (407), pretend the floor counts as a solution (405), or come from miscounting the cycle (409, 413).

💡Key takeaway

This AMC 8 problem only needs the Grade 5 pattern skill of "if I add the same rule each time, by how much does the total grow?" — that you already know!

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