Competition · AMC preparation · step 4 of 4

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

Write the balance equation

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

Bound the right jumps

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 405 right jumps

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 406 right jumps

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 407 right jumps

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

Confirm it is the smallest

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

Recognizing the "+3 right, +5 left" cycle that grows the total by 8 is the Grade 5 "generate and relate two numerical patterns" standard.

The working combination that uses the fewest left jumps also uses the fewest jumps in total, so no other combination can beat it.

▸ Why?

More left jumps always drag more right jumps along with them, so every other working combination ends up with a larger total.

▸ Why?

A left jump moves 3 pads backward, and to still finish exactly on the target that backward distance must be won back by adding right jumps, so the two counts rise together.

▸ Why?

Winning back distance you gave up is just undoing a backward jump with forward jumps, so extra left jumps force extra right jumps.

Inverse operationsSubtraction undoes addition and division undoes multiplication — they reverse each other.

▸ Why?

Where you finally land is the sum of every separate jump's move, so to keep that sum fixed at the target while the backward moves grow, the forward moves must grow too.

Whole is the sum of its partsBreak a thing into pieces with no gaps or overlaps and the pieces add back to the whole.

▸ Why?

There is a fixed smallest number of left jumps that still lets the right jumps of 5 land exactly on the target, and this combination is the one sitting at that smallest count.

▸ Why?

After the left jumps pull back some distance, the distance left over must split into whole jumps of 5, and the left-jump counts that allow this are spaced a fixed step apart, so the smallest one is well defined.

▸ Why?

Splitting the leftover distance into 5-pad jumps leaves a remainder, and the right jumps land exactly on the target only when that remainder is zero — that is, when the leftover is a whole multiple of 5.

Division with remainderDividing N into equal groups of size d gives N = q·d + r with 0 ≤ r < d; N is a multiple of d exactly when r = 0.

▸ Why?

The left-jump counts that work repeat on a fixed cycle, so once you find the smallest one the others are only full cycles later and therefore larger.

A cycle returns after a full periodA fixed repeating cycle of length p comes back to the same position every p steps.

5.OA.B.3Look For A Pattern

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!

Write the balance equation
Bound the right jumps
Try 405 right jumps
Try 406 right jumps
Try 407 right jumps
Confirm it is the smallest

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Write the balance equation

Bound the right jumps

Try 405 right jumps

Try 406 right jumps

Try 407 right jumps

Confirm it is the smallest

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