AMC 8 · 2023 · #18

Easy mode Grade 5

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Problem

Picture a long row of lily pads in a pond, all in a straight line.

Greta the grasshopper sits on one of the pads. From wherever she is, she can do exactly one of two moves:

Jump $5$ pads to the right, or
Jump $3$ pads to the left.

Every jump has to be one of those two. She can mix them in any order, as many times as she likes.

Greta wants to end up on the pad that is $2023$ pads to the right of where she started.

What is the smallest number of jumps she needs to get there?

$\textbf{(A) } 405 \qquad \textbf{(B) } 407 \qquad \textbf{(C) } 409 \qquad \textbf{(D) } 411 \qquad \textbf{(E) } 413$

Pick an answer.

(A)

405

(B)

407

(C)

409

(D)

411

(E)

413

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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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