AMC 8 · 2024 · #8
Easy mode Grade 4Problem
Imagine Taye has $2 on Monday.
Every day, Taye does exactly one of these two things to the money he had the day before:
- add $3 to it, or
- double it.
So on Tuesday his money depends on which choice he made, and the same is true again on Wednesday and Thursday.
By Thursday (three days after Monday), how many different dollar amounts could Taye possibly have?
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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