AMC 8 · 2024 · #8

Grade 4 arithmetic

Archetype Branching State Enumeration └ Sub-archetype Numeric Operation Tree

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

📏 Medium solution 💡 3 insights

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Problem

Taye starts Monday with

2</span>. Each day he chooses one of two actions: <span class="mk-c">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.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

There are only two choices per day for three days, so at most 2³ = 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.

1STEP 1

Apply +3 and double to Monday's $2 — 2 + 3 = 5 and 2 × 2 = 4 — so Tuesday's amounts are {4, 5}.

2 + 3 = 5, 2 × 2 = 4 → Tuesday ∈ {4, 5}

2STEP 2

Branch each Tuesday amount: 4→7,8 and 5→8,10; the raw list {7,8,8,10} repeats 8, so distinct Wednesday amounts are {7, 8, 10}.

4 &→ 4+3=7, 4× 2=8 ; 5 &→ 5+3=8, 5× 2=10 → Wednesday ∈ {7, 8, 10}

3STEP 3

Apply both actions to each Wednesday amount: 7→10,14; 8→11,16; 10→13,20, giving the Thursday list {10, 14, 11, 16, 13, 20} — six values.

7 &→ 7+3=10, 7× 2=14 ; 8 &→ 8+3=11, 8× 2=16 ; 10 &→ 10+3=13, 10× 2=20

4STEP 4

All six candidates are distinct, so the count of possible amounts is 6 — choice (D); 3, 4, 5, 7 don't match and are eliminated.

{10, 11, 13, 14, 16, 20} → #distinct = 6 → (D)

Answer

There are 2 × 2 × 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 → +3 and 5 × 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.

💡Key takeaway

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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