AMC 8 · 2024 · #8
Grade 4 arithmeticProblem
On Monday, Taye has 2$. Every day, he either gains3ordoublestheamountofmoneyhehadonthepreviousday.HowmanydifferentdollaramountscouldTayehaveonThursday,3$ days later?
Pick an answer.
AMC 8 2024 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.
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
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
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.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 ${4, 5}$.
💡 Adding within 20 and doubling small numbers are fluent Grade 2 mental-math operations.
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 $\{7, 8, 10\}$.
💡 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.
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.
💡 Carrying out the two operations within $100$ step by step on each prior amount is the multi-step, four-operation reasoning of Grade 3.
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.
💡 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.
2.OA.B.2 Apply the two possible actions to Monday's $\$2$ to list every Tuesday amount. T 3.OA.C.7 Branch each of Tuesday's two amounts again to get Wednesday. From $4$: $4 + 3 = 3.OA.D.8 Apply the two actions once more to each of Wednesday's three distinct amounts to 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.2Fluently 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.7Fluently 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.8Solve 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.5Generate 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!
⭐ This AMC 8 problem only needs the Grade 4 skill of generating number patterns by following a given rule that you already know!
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