AMC 8 · 2024 · #9

Grade 4 arithmetic
ratio-proportionmultiplespattern-recognition pattern-recognitionsystematic-enumeration ↑ Prerequisites: multi-digit-arithmeticmultiples
📏 Medium solution 💡 3 insights
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Problem
Maria's marbles are all red, green, or blue. The number of red marbles is half the number of green marbles, and the number of blue marbles is twice the number of green marbles. Which of the choices 24, 25, 26, 27, 28 could be the total number of marbles?

Pick an answer.

(A)
24
(B)
25
(C)
26
(D)
27
(E)
28

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

How to solve
Strategy Guess and Check

Once you notice G must be even, the most natural move is Tool #6 (Guess and Check): try the smallest even green counts G = 2, 4, 6, 8, … and just build the collection. After a few tries, Tool #5 (Look for a Pattern) reveals the totals are 7, 14, 21, 28, … — multiples of 7. Since the problem is multiple choice, Tool #3 (Eliminate Possibilities) finishes the job by keeping only the choice that is a multiple of 7. This avoids algebra entirely; we ‘build’ the answer with concrete numbers.

1STEP 1

Red is half of green, so green must be even; the smallest even case G = 2 gives R = 1, B = 4.

G = 2 → R = 12\frac{1}{2}· 2 = 1, B = 2· 2 = 4
2STEP 2

Adding the three counts gives the smallest total: 1 + 2 + 4 = 7.

T = 1 + 2 + 4 = 7
3STEP 3

Trying G = 4, 6, 8 gives totals 14, 21, 28 — they rise by 7 each time, so every total is a multiple of 7.

T ∈ {7, 14, 21, 28, 35, …} = {7k : k ≥ 1}
4STEP 4

Among the choices, only 28 = 7 × 4 is a multiple of 7, and (R,G,B) = (4,8,16) confirms it works.

28 = 7 × 4 → (E) 28
Answer
28
Plug the answer 28 back into the original conditions. Green 8, red 4 (= 8 ÷ 2, half of green ✓), blue 16 (= 8 × 2, twice green ✓), total 4 + 8 + 16 = 28 ✓. All three counts are whole numbers, so every condition is satisfied. Also, one ‘bundle’ of (1 red, 2 green, 4 blue) is 7 marbles, and the total is just k bundles, so only multiples of 7 are possible — confirming why 24, 25, 26, 27 cannot occur.
💡Key takeaway

This AMC 8 problem only needs Grade 4 multiplicative comparison and multiples of 7 you already know!