🇺🇸 AMC 8 ⇄ switch contest
2012 AMC 8

Problem 19

Problem 19 · 2012 AMC 8 Easy
Algebra & Patterns all-but-trick

In a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. How many marbles are in the jar?

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Answer: C — 9 marbles.
Show hints
Hint 1 of 2
Flip each statement around. "All but 6 are red" isn't about red — it says the non-red marbles (green + blue) number 6. Rewrite all three the same way.
Still stuck? Show hint 2 →
Hint 2 of 2
Now add your three equations and watch what happens: each color shows up in exactly two of them. The technique is add-them-all and exploit the symmetry.
Show solution
Approach: rephrase as two-color sums, then add
  1. Reverse each clue into a count of the other two colors: green + blue = 6, red + blue = 8, red + green = 4.
  2. Add all three. Every color appears in exactly two equations, so the left side is 2(red + green + blue): 2 × Total = 6 + 8 + 4 = 18.
  3. Therefore Total = 18 / 2 = 9 marbles.
  4. Why the ÷2: summing "all but" statements double-counts every marble, so the total is half the sum of the three given numbers — a handy shortcut for any "all but" puzzle.
Another way — find each color, then total:
  1. From green + blue = 6 and red + blue = 8, subtract: red − green = 2. Combine with red + green = 4 to get red = 3, green = 1.
  2. Then blue = 6 − green = 5. Total = 3 + 1 + 5 = 9, matching the shortcut.
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