AMC 8 · 2017 · #17

Grade 4 algebra

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

linear-equations-one-varsystems-of-equationsmulti-digit-arithmetic convert-to-algebraguess-and-check ↑ Prerequisites: multi-digit-arithmeticlinear-equations-one-var

📏 Medium solution 💡 3 insights

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Problem

Starting with some gold coins and some empty treasure chests, I tried to put $9$ gold coins in each treasure chest, but that left $2$ treasure chests empty. So instead I put $6$ gold coins in each treasure chest, but then I had $3$ gold coins left over. How many gold coins did I have?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2017 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: You have a pile of gold coins and a row of empty treasure chests. If you try to put $9$ coins in every chest, you run out of coins before filling everyone — $2$ chests are left empty. If instead you put $6$ coins in every chest, every chest is full and you still have $3$ coins left in your hand. How many gold coins do you have?

Givens: Plan A: $9$ coins per chest fills only the first $(C-2)$ chests and uses every coin, where $C$ is the total number of chests; Plan B: $6$ coins per chest fills all $C$ chests and leaves $3$ coins left over; Both plans use the SAME pile of coins and the SAME number of chests; Answer choices: (A) $9$, (B) $27$, (C) $45$, (D) $63$, (E) $81$

Unknowns: The total number of gold coins, $G$

Understand

Plan

Primary tool: #3 Eliminate Possibilities

Secondary: #6 Guess and Check

It is a multiple-choice AMC problem with only five candidate values for $G$, and each candidate is easy to plug back into the story. Tool #3 (Eliminate Possibilities) is the natural first move on any AMC multiple-choice problem — for every candidate $G$ we can compute the number of chests two different ways (once from Plan A and once from Plan B) and keep only the candidate where the two answers agree. Tool #6 (Guess and Check) is what powers each individual test: guess a $G$, compute, and see if the two pictures line up. This is faster and more elementary than setting up a system of equations (Tool #13).

Execute — Answer: C

#3 Eliminate Possibilities 4.OA.A.3 Step 1

Translate the two stories into arithmetic.
Plan A uses $C-2$ chests of $9$ coins, so $G = 9 \times (C-2)$, which means $C = \dfrac{G}{9} + 2$.
Plan B uses $C$ chests of $6$ coins plus $3$ leftover, so $G = 6C + 3$, which means $C = \dfrac{G-3}{6}$.

$$C_{A} = \dfrac{G}{9} + 2 \qquad C_{B} = \dfrac{G-3}{6}$$

💡 Reading a multi-step word problem and turning each sentence into a small calculation is a Grade 4 multi-step word-problem skill.

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

Test candidate (A) $G = 9$.
Plan A: $C_A = 9/9 + 2 = 3$ chests.
Plan B: $C_B = (9-3)/6 = 1$ chest.
$3 \ne 1$, so the same pile of coins would need two different numbers of chests — impossible.
Eliminate (A).

$$C_A = 3,\; C_B = 1 \;\Rightarrow\; \text{mismatch}$$

💡 All four checks are just dividing and adding within $100$, which is Grade 3 multiplication/division fluency.

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

Test candidate (B) $G = 27$.
Plan A: $C_A = 27/9 + 2 = 5$ chests.
Plan B: $C_B = (27-3)/6 = 4$ chests.
Still a mismatch, so eliminate (B).

$$C_A = 5,\; C_B = 4 \;\Rightarrow\; \text{mismatch}$$

💡 Same Grade 3 division facts ($27 \div 9$, $24 \div 6$) used inside the elimination loop.

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

Test candidate (C) $G = 45$.
Plan A: $C_A = 45/9 + 2 = 5 + 2 = 7$ chests.
Plan B: $C_B = (45-3)/6 = 42/6 = 7$ chests.
Both pictures agree on $7$ chests, so (C) survives the test.

$$C_A = 7 = C_B \;\Rightarrow\; G = 45 \;\checkmark$$

💡 Finding the unknown number of chests that makes both equations true is exactly Grade 3 "determine the unknown whole number in a multiplication or division equation."

#3 Eliminate Possibilities 4.OA.A.3 Step 5

For completeness, finish eliminating the rest.
(D) $G = 63$: $C_A = 63/9 + 2 = 9$, $C_B = (63-3)/6 = 10$ — mismatch.
(E) $G = 81$: $C_A = 81/9 + 2 = 11$, $C_B = (81-3)/6 = 13$ — mismatch.
Only (C) survives, so the answer is $G = 45$, choice (C).

$$45 = 9 \times (7-2) = 9 \times 5,\qquad 45 = 6 \times 7 + 3 \;\Rightarrow\; \textbf{(C)}$$

💡 Knocking out the other choices and finishing the multi-step verification is the final Grade 4 word-problem move.

[1] #3 4.OA.A.3 Translate the two stories into arithmetic. Plan A uses $C-2$ chests of $9$ coins

[2] #6 3.OA.C.7 Test candidate (A) $G = 9$. Plan A: $C_A = 9/9 + 2 = 3$ chests. Plan B: $C_B = (

[3] #6 3.OA.C.7 Test candidate (B) $G = 27$. Plan A: $C_A = 27/9 + 2 = 5$ chests. Plan B: $C_B =

[4] #6 3.OA.A.4 Test candidate (C) $G = 45$. Plan A: $C_A = 45/9 + 2 = 5 + 2 = 7$ chests. Plan B

[5] #3 4.OA.A.3 For completeness, finish eliminating the rest. (D) $G = 63$: $C_A = 63/9 + 2 = 9

Review

Reasonableness: With $G = 45$ and $C = 7$: Plan A fills $5$ chests with $9$ coins each ($5 \times 9 = 45$, all coins used, $2$ chests left empty — matches the problem). Plan B fills all $7$ chests with $6$ coins ($7 \times 6 = 42$) and leaves $45 - 42 = 3$ coins over — also matches. Both stories are satisfied by the same numbers, so $45$ is consistent.

Alternative: Tool #13 (Convert to Algebra) also works: set $9(C-2) = 6C + 3$, expand to $9C - 18 = 6C + 3$, get $3C = 21$, so $C = 7$ and $G = 6 \times 7 + 3 = 45$. The algebra is clean, but for an AMC multiple-choice problem with only five small candidates, plugging in the choices (Tool #3 + Tool #6) is just as fast and uses only Grade 3-4 arithmetic.

CCSS standards used (min grade 4)

3.OA.C.7 Fluently multiply and divide within 100 (Computing $G/9$ and $(G-3)/6$ for each candidate $G \in \{9, 27, 45, 63, 81\}$ — all divisions and multiplications stay within $100$.)
3.OA.A.4 Determine unknown whole number in multiplication or division equation (Finding the chest count $C$ that makes both $G = 9(C-2)$ and $G = 6C + 3$ true at the same time for the surviving candidate $G = 45$.)
4.OA.A.3 Solve multi-step word problems using four operations with whole numbers (Reading the two coin-distribution scenarios, translating each into a small whole-number equation, and verifying a candidate against both at once.)

⭐ This AMC 8 problem only needs Grade 4 multi-step word-problem skills you already know — just check each answer choice in both stories and keep the one that fits!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 4)

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