Problem 9 · 2024 AMC 8
Medium
Number Theory
divisibilitysubstitution
All of the marbles in Maria's collection are red, green, or blue. Maria has half as many red marbles as green marbles and twice as many blue marbles as green marbles. Which of the following could be the total number of marbles in Maria's collection?
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Answer: E — 28 marbles.
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Hint 1 of 2
The three colors lock together in fixed ratios, so the total can't be just any number. Name the smallest pile with a variable and the others follow — what does their sum have to be a multiple of?
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Hint 2 of 2
Technique: pick the color that makes the others whole. Red is smallest, so let r = red, green = 2r, blue = 4r. Total 7r must be a multiple of 7.
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Approach: fix the ratios into one variable, find the hidden multiple
- Choose the variable to dodge fractions. "Half as many red as green" makes green double the red, so let r = red (the smallest). Then green = 2r, and blue = twice green = 4r.
- Total = r + 2r + 4r = 7r — whatever r is, the total is a multiple of 7.
- Only 28 = 7 × 4 among the choices is a multiple of 7. This transfers: when quantities are tied by ratios, the total is always a fixed multiple, so the answer must be divisible by the sum of the ratio parts (here 1 + 2 + 4 = 7) — you can often skip straight to that divisibility test.
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