AMC 8 · 2024 · #8

Grade 4 arithmetic

pattern-recognitionsystematic-enumeration tree-enumerationsystematic-enumeration ↑ Prerequisites: multi-digit-arithmeticmental-arithmetic

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Problem

On Monday, Taye has $2$. Every day, he either gains$ 3 $or doubles the amount of money he had on the previous day. How many different dollar amounts could Taye have on Thursday,$ 3$ days later?

$\textbf{(A) } 3\qquad\textbf{(B) } 4\qquad\textbf{(C) } 5\qquad\textbf{(D) } 6\qquad\textbf{(E) } 7$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: Taye starts Monday with $\$2$. Each day he chooses one of two actions: add $\$3$, or double his money. We need to count how many different dollar amounts he can possibly have on Thursday, which is $3$ days (and so $3$ actions) after Monday.

Givens: Monday starting amount: $\$2$; Each day, one of two actions: $+3$ or $\times 2$; Monday $\to$ Tuesday $\to$ Wednesday $\to$ Thursday: exactly 3 actions; Five answer choices: (A) 3, (B) 4, (C) 5, (D) 6, (E) 7

Unknowns: The number of distinct dollar amounts Taye can have on Thursday

Understand

Plan

Primary tool: #2 Make a Systematic List

Secondary: #1 Draw a Diagram, #3 Eliminate Possibilities

There are only two choices per day for three days, so at most $2^3 = 8$ possible paths — small enough to write down every single one (Tool #2). Drawing the day-by-day possibilities as a branching tree (Tool #1) makes duplicate amounts (different paths, same total) impossible to miss. At the end we cross out duplicates and match against the answer choices (Tool #3). No algebra is needed.

Execute — Answer: D

#2 Make a Systematic List 2.OA.B.2 Step 1

Apply the two possible actions to Monday's $\$2$ to list every Tuesday amount. The two children of the root are $2 + 3 = 5$ and $2 \times 2 = 4$. So Tuesday's possible amounts are ${4, 5}$.

$$2 + 3 = 5,\quad 2 \times 2 = 4 \;\Rightarrow\; \text{Tuesday} \in \{4,\,5\}$$

💡 Adding within 20 and doubling small numbers are fluent Grade 2 mental-math operations.

#1 Draw a Diagram 3.OA.C.7 Step 2

Branch each of Tuesday's two amounts again to get Wednesday.
From $4$: $4 + 3 = 7$ and $4 \times 2 = 8$.
From $5$: $5 + 3 = 8$ and $5 \times 2 = 10$.
The raw list is $\{7, 8, 8, 10\}$ — note that $8$ shows up twice (two different paths collide), so the set of **distinct** Wednesday amounts is $\{7, 8, 10\}$.

$$\begin{aligned} 4 &\to 4+3=7,\ 4\times 2=8 \\ 5 &\to 5+3=8,\ 5\times 2=10 \end{aligned} \;\Rightarrow\; \text{Wednesday} \in \{7,\,8,\,10\}$$

💡 A branching tree diagram makes the two paths joining at $8$ obvious. Single-digit doublings ($4\times 2$, $5 \times 2$) are part of Grade 3 multiplication fluency.

#2 Make a Systematic List 3.OA.D.8 Step 3

Apply the two actions once more to each of Wednesday's three distinct amounts to enumerate every Thursday possibility.
From $7$: $10, 14$.
From $8$: $11, 16$.
From $10$: $13, 20$.
The full Thursday list is $\{10, 14, 11, 16, 13, 20\}$ — six values in total.

$$\begin{aligned} 7 &\to 7+3=10,\ 7\times 2=14 \\ 8 &\to 8+3=11,\ 8\times 2=16 \\ 10 &\to 10+3=13,\ 10\times 2=20 \end{aligned}$$

💡 Carrying out the two operations within $100$ step by step on each prior amount is the multi-step, four-operation reasoning of Grade 3.

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

Check the six Thursday candidates $\{10, 14, 11, 16, 13, 20\}$ for duplicates.
Sorted: $10, 11, 13, 14, 16, 20$ — all six are distinct.
The number of possible distinct amounts is $6$, which matches answer choice (D).
The choices $3, 4, 5, 7$ disagree with our explicit list and are eliminated.

$$\{10,\,11,\,13,\,14,\,16,\,20\} \;\Rightarrow\; \#\text{distinct} = 6 \;\Rightarrow\; \textbf{(D)}$$

💡 Generating the terms of a sequence by repeatedly applying a given rule ($+3$ or $\times 2$) and counting distinct outputs is exactly the Grade 4 'generate a pattern following a rule' standard.

[1] #2 2.OA.B.2 Apply the two possible actions to Monday's $\$2$ to list every Tuesday amount. T

[2] #1 3.OA.C.7 Branch each of Tuesday's two amounts again to get Wednesday. From $4$: $4 + 3 =

[3] #2 3.OA.D.8 Apply the two actions once more to each of Wednesday's three distinct amounts to

[4] #3 4.OA.C.5 Check the six Thursday candidates $\{10, 14, 11, 16, 13, 20\}$ for duplicates. S

Review

Reasonableness: There are $2 \times 2 \times 2 = 8$ possible action sequences in total. If no two sequences ever produced the same dollar amount, the answer would be $8$. We did find one collision on Wednesday (both $4 \to +3$ and $5 \times 2$ give $\$8$), which propagates to two redundant Thursday values, knocking the count down to $8 - 2 = 6$. So $6$ is consistent with the path count, and the answer (D) is correct.

Alternative: Tool #1 (Draw a Diagram) on its own works just as well: draw a binary tree with $\$2$ at the root, branching to $+3$ on the left and $\times 2$ on the right at every level, and read the eight leaves. Visually you can see which leaves merge. The systematic list and the tree are two representations of the same casework — pick whichever feels cleaner.

CCSS standards used (min grade 4)

2.OA.B.2 Fluently add and subtract within 20 using mental strategies (Computing the first-day transitions like $2 + 3 = 5$ and small doublings such as $2 \times 2 = 4$ from Monday to Tuesday.)
3.OA.C.7 Fluently multiply and divide within 100 (Doubling Tuesday and Wednesday amounts ($4\times 2, 5\times 2, 7\times 2, 8\times 2, 10 \times 2$) using single-digit multiplication facts.)
3.OA.D.8 Solve two-step word problems using four operations within 100 (Applying the two operations again to Wednesday's amounts to build every Thursday candidate in a multi-step way.)
4.OA.C.5 Generate a number or shape pattern following a given rule (Repeatedly applying the rule '$+3$ or $\times 2$' for three days to generate every possible Thursday amount and counting distinct results.)

⭐ This AMC 8 problem only needs the Grade 4 skill of generating number patterns by following a given rule that you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 4)

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