Problem 9 · 2017 AMC 8
Medium
Number Theory
divisibilitycasework
All of Marcy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles that Marcy could have?
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Answer: D — 4 yellow marbles.
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Hint 1 of 2
You can't have a third of a marble. Since 1/3 and 1/4 of the total are both whole counts, the total is locked to multiples of 12 — you only ever test 12, 24, 36, …
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Hint 2 of 2
Try the smallest legal total first. If blue + red + green already exceeds it, that total is impossible and you bump up to the next multiple of 12.
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Approach: total must be a multiple of lcm(3, 4) = 12
- Blue is 1/3 and red is 1/4 of the total, and counts are whole numbers, so the total must be divisible by both 3 and 4 — i.e. a multiple of 12. That restriction is the whole problem.
- Try 12: blue = 4, red = 3, plus 6 green = 13 marbles — already more than 12. Impossible, so jump up.
- Try 24: blue = 8, red = 6, green = 6, leaving yellow = 24 − 8 − 6 − 6 = 4. This works and is the smallest valid total.
- Why smallest total = smallest yellow here: once green (6) is fixed, a larger total only adds more blue+red+yellow, so the first total that fits gives the fewest yellow.
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