AMC 8 · 2017 · #9

Easy mode Grade 4

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Problem

Marcy has a pile of marbles. Every marble is one of four colors: blue, red, green, or yellow.

Exactly one third of her marbles are blue. Exactly one fourth of her marbles are red. And $6$ of her marbles are green.

The rest of her marbles are yellow. We want to make that "rest" as small as possible.

What is the smallest number of yellow marbles Marcy could have?

$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }5$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: Marcy's marbles come in four colors: blue, red, green, yellow. Exactly $\tfrac{1}{3}$ of them are blue, exactly $\tfrac{1}{4}$ are red, and exactly $6$ are green. We want the smallest possible number of yellow marbles she could have.

Givens: Blue marbles $= \tfrac{1}{3}$ of the total; Red marbles $= \tfrac{1}{4}$ of the total; Green marbles $= 6$; All other marbles are yellow; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$

Unknowns: The smallest possible number of yellow marbles

Understand

Plan

Primary tool: #9 Solve an Easier Related Problem

Secondary: #6 Guess and Check, #3 Eliminate Possibilities

Instead of writing an equation with an unknown total, we use Tool #9: replace the abstract "total" with the smallest concrete totals that work, and check each one. Because $\tfrac{1}{3}$ and $\tfrac{1}{4}$ of the total must both be whole numbers, the total has to be a multiple of $3$ and of $4$ — so the candidates are $12, 24, 36, \ldots$. Tool #6 (Guess and Check) walks through these smallest totals in order; the first total that leaves a non-negative number of yellow marbles is the answer. Tool #3 (Eliminate) keeps us honest: once we find $Y = 4$, we can confirm that the smaller answer choices $(1, 2, 3)$ are impossible.

Execute — Answer: D

#9 Solve an Easier Related Problem 4.OA.B.4 Step 1

Decide which totals are even allowed.
Since $\tfrac{1}{3}$ of the total must be a whole number of blue marbles, the total is a multiple of $3$.
Since $\tfrac{1}{4}$ of the total must be a whole number of red marbles, the total is also a multiple of $4$.
The numbers that are multiples of both $3$ and $4$ are $12, 24, 36, 48, \ldots$.

$$\text{Possible totals} \in \{12,\, 24,\, 36,\, \ldots\}$$

💡 Listing common multiples of $3$ and $4$ is exactly the Grade 4 "multiples" skill — no algebra needed.

#6 Guess and Check 3.OA.A.3 Step 2

Try the smallest candidate, total $= 12$.
Then blue $= \tfrac{1}{3}\cdot 12 = 4$ and red $= \tfrac{1}{4}\cdot 12 = 3$.
Add the $6$ green: $4 + 3 + 6 = 13$.
But the total was supposed to be $12$, so we'd already have $1$ marble too many before counting any yellow.
Impossible — throw this candidate out.

$$4 + 3 + 6 = 13 > 12 \;\Rightarrow\; \text{no room for yellow}$$

💡 Multiplying and adding small whole numbers is Grade 3 — perfect for plugging in a guess and checking the fit.

#6 Guess and Check 3.OA.D.8 Step 3

Try the next candidate, total $= 24$.
Then blue $= \tfrac{1}{3}\cdot 24 = 8$ and red $= \tfrac{1}{4}\cdot 24 = 6$.
With $6$ green, the colored marbles other than yellow add up to $8 + 6 + 6 = 20$, leaving $24 - 20 = 4$ yellow marbles.
That's a valid, non-negative whole number, so $Y = 4$ works.

$$8 + 6 + 6 = 20,\;\; 24 - 20 = 4 \text{ yellow}$$

💡 Multi-step word problems with the four operations within $100$ are Grade 3 — no fractions to add, just careful arithmetic.

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

Confirm that $4$ really is the smallest.
Any total smaller than $24$ that satisfies the divisibility rule is just $12$, and we already showed $12$ leaves no room for yellow.
Larger totals like $36$ would give even more yellow marbles ($\tfrac{1}{3}\cdot 36 + \tfrac{1}{4}\cdot 36 + 6 = 12 + 9 + 6 = 27$, so yellow $= 36 - 27 = 9$).
So the minimum is exactly $4$.

$$T=12 \to \text{invalid};\;\; T=24 \to Y=4;\;\; T=36 \to Y=9;\;\; \ldots$$

💡 Sweeping through valid totals in order and stopping at the first success is the Grade 4 "multi-step word problem with reasoning about results" move.

#3 Eliminate Possibilities 2.NBT.B.5 Step 5

Match the answer to the choice list.
The smallest possible number of yellow marbles is $4$, which is choice (D).

$$Y_{\min} = 4 \;\Rightarrow\; \textbf{(D)}$$

💡 Reading off the answer choice that matches $4$ uses only Grade 2 number recognition.

[1] #9 4.OA.B.4 Decide which totals are even allowed. Since $\tfrac{1}{3}$ of the total must be

[2] #6 3.OA.A.3 Try the smallest candidate, total $= 12$. Then blue $= \tfrac{1}{3}\cdot 12 = 4$

[3] #6 3.OA.D.8 Try the next candidate, total $= 24$. Then blue $= \tfrac{1}{3}\cdot 24 = 8$ and

[4] #3 4.OA.A.3 Confirm that $4$ really is the smallest. Any total smaller than $24$ that satisf

[5] #3 2.NBT.B.5 Match the answer to the choice list. The smallest possible number of yellow marb

Review

Reasonableness: Sanity-check with $T = 24$: $8$ blue $+$ $6$ red $+$ $6$ green $+$ $4$ yellow $= 24$. Fractions check out: $\tfrac{8}{24} = \tfrac{1}{3}$ blue and $\tfrac{6}{24} = \tfrac{1}{4}$ red, exactly as the problem says. Yellow is the smallest of all four counts ($4 < 6 < 6 \le 8$), which fits the question's hint that yellow is the leftover, not the dominant, color. Answer $4$ is also one of the listed choices, so we are not off by a factor.

Alternative: Tool #16 (Change Focus / Complement) gives a quick formula-free check. Blue plus red eat up $\tfrac{1}{3} + \tfrac{1}{4} = \tfrac{7}{12}$ of the marbles, so green plus yellow are the remaining $\tfrac{5}{12}$. If the total were $12$, green $+$ yellow would have to equal $5$, but green alone is $6$, which is too many. The next allowed total is $24$, where green $+$ yellow $= \tfrac{5}{12}\cdot 24 = 10$, leaving yellow $= 10 - 6 = 4$. Same answer.

CCSS standards used (min grade 4)

2.NBT.B.5 Fluently add and subtract within 100 (Adding the color counts ($8+6+6=20$) and subtracting from the total ($24-20=4$) to find the number of yellow marbles.)
3.OA.A.3 Solve multiplication and division word problems within 100 (Computing fractional parts of small totals like $\tfrac{1}{3}\cdot 12 = 4$ and $\tfrac{1}{4}\cdot 12 = 3$ as simple division/multiplication.)
3.OA.D.8 Solve two-step word problems using four operations within 100 (Combining several operations (find blue, find red, add green, subtract from total) to test whether each candidate total works.)
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Recognizing that the total must be a common multiple of $3$ and $4$ and listing the candidates $12, 24, 36, \ldots$.)
4.OA.A.3 Solve multi-step word problems using four operations with whole numbers (Reasoning across multiple candidate totals — eliminating $T=12$, accepting $T=24$ — to identify the smallest valid yellow count.)

⭐ This AMC 8 problem only needs Grade 4 thinking about multiples of $3$ and $4$ that you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 4)

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