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

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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?

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.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: 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?

Givens: Two possible moves per jump: $+5$ (right) or $-3$ (left); Target net displacement: exactly $+2023$ pads from the start; Answer choices: (A) $405$, (B) $407$, (C) $409$, (D) $411$, (E) $413$

Unknowns: The minimum total number of jumps (right jumps $+$ left jumps) Greta needs

Understand

Plan

Primary tool: #6 Guess and Check

Secondary: #9 Solve an Easier Related Problem, #5 Look for a Pattern

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 $R = 405, 406, 407, \dots$ 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 $3$s?" Tool #5 (Pattern) confirms the divisibility-by-$3$ rule on the overshoot.

Execute — Answer: D

#9 Solve an Easier Related Problem 3.OA.A.3 Step 1

Translate the situation into a balance equation.
Each right jump adds $5$ and each left jump subtracts $3$.
To finish at $+2023$, the total of all moves must be $2023$:

$$5R - 3L = 2023, \quad R \ge 0,\; L \ge 0$$

💡 Writing a multiplication-and-subtraction equation for a real-world count is the Grade 3 "multiplication and division word problems" skill.

#6 Guess and Check 3.OA.C.7 Step 2

Find a lower bound for $R$.
Since each left jump only subtracts from her position, she needs at least enough right jumps to reach $2023$ on their own.
Because $5 \times 404 = 2020 < 2023$, she needs $R \ge 405$.

$$5 \times 404 = 2020 < 2023 < 2025 = 5 \times 405 \;\Rightarrow\; R \ge 405$$

💡 Comparing $5 \times 404$ and $5 \times 405$ to $2023$ is Grade 3 multiplication-table fluency.

#6 Guess and Check 4.OA.B.4 Step 3

Try $R = 405$.
Then $5R = 2025$, so the overshoot is $2025 - 2023 = 2$.
We would need $3L = 2$, but $2$ is not a multiple of $3$, so $L$ would not be a whole number.
Reject.

$$5(405) - 3L = 2023 \;\Rightarrow\; 3L = 2 \;\Rightarrow\;\text{no whole-number } L$$

💡 Checking whether $2$ is a multiple of $3$ is the Grade 4 "multiples and factor pairs" idea.

#6 Guess and Check 4.OA.B.4 Step 4

Try $R = 406$.
Then $5R = 2030$, overshoot $= 7$.
Is $7$ a multiple of $3$?
No ($3 \times 2 = 6,\; 3 \times 3 = 9$).
Reject.

$$5(406) - 3L = 2023 \;\Rightarrow\; 3L = 7 \;\Rightarrow\;\text{no whole-number } L$$

💡 Again testing a small number against the multiples of $3$ — same Grade 4 multiple-of skill.

#6 Guess and Check 4.NBT.B.6 Step 5

Try $R = 407$.
Then $5R = 2035$, overshoot $= 12$.
Now $12 = 3 \times 4$, so $L = 4$ works.
Total jumps so far: $R + L = 407 + 4 = 411$.

$$5(407) - 3L = 2023 \;\Rightarrow\; 3L = 12 \;\Rightarrow\; L = 4 \;\Rightarrow\; R + L = 411$$

💡 Solving $5R = 2035$ to get $R = 407$ and $3L = 12$ to get $L = 4$ is exactly Grade 4 "whole-number quotients with multi-digit dividends."

#5 Look for a Pattern 5.OA.B.3 Step 6

Confirm this is the minimum by spotting the pattern: every time we add $3$ more right jumps, the overshoot grows by $5 \times 3 = 15$, but to absorb $15$ with left jumps requires $L$ to grow by $5$, so the total $R + L$ grows by $3 + 5 = 8$.
Starting from the first valid solution, any other valid $(R, L)$ uses MORE jumps.
So $R = 407,\; L = 4,\; \text{total} = 411$ is the minimum, matching choice (D).

$$R + L = 411 \;\Rightarrow\;\textbf{(D)}\ 411$$

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

[1] #9 3.OA.A.3 Translate the situation into a balance equation. Each right jump adds $5$ and ea

[2] #6 3.OA.C.7 Find a lower bound for $R$. Since each left jump only subtracts from her positio

[3] #6 4.OA.B.4 Try $R = 405$. Then $5R = 2025$, so the overshoot is $2025 - 2023 = 2$. We would

[4] #6 4.OA.B.4 Try $R = 406$. Then $5R = 2030$, overshoot $= 7$. Is $7$ a multiple of $3$? No (

[5] #6 4.NBT.B.6 Try $R = 407$. Then $5R = 2035$, overshoot $= 12$. Now $12 = 3 \times 4$, so $L

[6] #5 5.OA.B.3 Confirm this is the minimum by spotting the pattern: every time we add $3$ more

Review

Reasonableness: If Greta only jumped right, she would need $2023/5 = 404.6$ 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 ($409, 413$).

Alternative: Tool #3 (Eliminate Possibilities) on the answer choices. Note that $R + L \equiv R - L \pmod{2}$ is not very useful here, but we can use divisibility: from $5R - 3L = 2023$ and $R + L = J$, we get $L = \dfrac{5J - 2023}{8}$, so $J$ must make $5J - 2023$ a multiple of $8$. Test each choice: $5(405)-2023=2$ (no); $5(407)-2023=12$ (no); $5(409)-2023=22$ (no); $5(411)-2023=32 = 8 \times 4$ (yes, so $L = 4$); $5(413)-2023=42$ (no). Only $411$ survives, confirming (D).

CCSS standards used (min grade 5)

3.OA.A.3 Solve multiplication and division word problems within 100 (Translating "each right jump is $+5$, each left jump is $-3$, end at $+2023$" into the balance equation $5R - 3L = 2023$.)
3.OA.C.7 Fluently multiply and divide within 100 (Computing products like $5 \times 404 = 2020$ and $5 \times 405 = 2025$ to bracket $2023$ and find a lower bound on $R$.)
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Checking whether the overshoot ($2, 7, 12, \dots$) is a multiple of $3$, since $3L$ must hit it exactly for $L$ to be a whole number.)
4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends (Dividing $2035 \div 5 = 407$ to find $R$ and $12 \div 3 = 4$ to find $L$ from the valid case.)
5.OA.B.3 Generate two numerical patterns using two given rules and identify relationships (Spotting that each "add $3$ right jumps, add $5$ left jumps" cycle increases the total by $8$, so the first valid $(R, L)$ is already the minimum.)

⭐ 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 5)

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