Problem 8 · 2011 AMC 8
Easy
Counting & Probability
enumerate-sums
Bag A has three chips labeled 1, 3, and 5. Bag B has three chips labeled 2, 4, and 6. If one chip is drawn from each bag, how many different values are possible for the sum of the two numbers on the chips?
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Answer: B — 5 different values.
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Hint 1 of 2
Bag A is all odd, Bag B is all even, so every sum is odd + even = odd. That instantly rules out even totals — you only need to count which odd numbers are reachable.
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Hint 2 of 2
Find the smallest and largest possible sum; every odd number in between turns out to be reachable, so just count the odds in that range.
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Approach: all sums are odd — count the odds from smallest to largest
- Smallest sum: 1 + 2 = 3. Largest: 5 + 6 = 11. Every sum is odd, so the only candidates are 3, 5, 7, 9, 11.
- Each of those is achievable (e.g. 5 = 1+4 or 3+2, 7 = 1+6 = 3+4 = 5+2, etc.), so all 5 occur.
- That's 5 distinct values.
- Worth keeping: spotting that odd + even is always odd cuts the candidate list in half before you compute anything — check the parity of a sum first.
Another way — grid of all 9 sums:
- The nine sums are 3, 5, 7 / 5, 7, 9 / 7, 9, 11.
- Deduping leaves {3, 5, 7, 9, 11} = 5 values.
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