Problem 24 · 2007 AMC 8
Medium
Counting & Probability
divisibility-by-3subset-counting
A bag contains four pieces of paper, each labeled with one of the digits 1, 2, 3, or 4, with no repeats. Three of these pieces are drawn, one at a time without replacement, to construct a three-digit number. What is the probability that the three-digit number is a multiple of 3?
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Answer: C — 1/2.
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Hint 1 of 2
Divisibility by 3 cares about the digit-sum, and a sum doesn't notice the order — so forget the three-digit arrangements entirely and just ask which trio of digits got drawn.
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Hint 2 of 2
Divisible by 3 ⇔ digit-sum divisible by 3. Since order can't change a sum, this is really a 'which 3 did I pick' question, not a 'which number' question.
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Approach: drawing 3 of 4 = leaving 1 out
- Order is irrelevant for divisibility by 3, so picking 3 of the 4 digits is the same as choosing which single digit to leave behind.
- All four digits sum to 1+2+3+4 = 10. Leaving out d makes the drawn sum 10 − d, which is a multiple of 3 exactly when d = 1 (sum 9) or d = 4 (sum 6).
- So 2 of the 4 equally likely 'leftover' digits work: probability = 2/4 = 1/2.
- Slick reframing: 'choose 3 from 4' is always 'discard 1 from 4' — turning a messy selection into one easy choice is a counting habit worth keeping.
Another way — list the four subsets directly:
- The size-3 subsets of {1,2,3,4} are {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4} with sums 6, 7, 8, 9.
- Multiples of 3: sums 6 and 9 — 2 of the 4 subsets. Probability 2/4 = 1/2.
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