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AMC 8 · 2017 · #17

Easy mode Grade 4
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Problem

Imagine a stack of gold coins and a row of empty treasure chests.

First try: I tried to put 99 coins into each chest. When I ran out of coins, 22 chests were still empty.

Second try: I started over and put 66 coins into each chest instead. This time every chest got coins, but I had 33 coins left over at the end.

The number of chests is the same in both tries. The number of coins is the same in both tries.

How many gold coins did I have?

(A) 9(B) 27(C) 45(D) 63(E) 81\textbf{(A) }9\qquad\textbf{(B) }27\qquad\textbf{(C) }45\qquad\textbf{(D) }63\qquad\textbf{(E) }81

Pick an answer.

(A)
9
(B)
27
(C)
45
(D)
63
(E)
81
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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

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) $81$

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!

⭐ 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!

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