Problem 21 · 2023 AMC 8
Hard
Counting & Probability
careful-countingcaseworksum-constraint
Alina writes the numbers 1, 2, …, 9 on separate cards, one number per card. She wishes to divide the cards into 3 groups of 3 cards so that the sum of the numbers in each group will be the same. In how many ways can this be done?
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Answer: C — 2 ways.
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Hint 1 of 2
Find the target sum first — that single number tells you a lot. In particular, where can the three largest cards possibly sit?
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Hint 2 of 2
Each group sums to 15, so 7, 8, 9 can't share a group (7+8 already overshoots room) — they're spread one per group. Same for 1, 2, 3. Now the only freedom left is where 5 lands; case-split on that.
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Approach: fix the totals, then place the extreme numbers
- The total 1+2+…+9 = 45 splits into three equal groups, so each must sum to 15. That target instantly pins the big and small cards.
- 7, 8, 9 must each land in a different group (any two of them already total 15 or more, leaving no room for a positive third). By the same squeeze, 1, 2, 3 spread out one per group too.
- So the only real choice is 5's partner-pair, which must sum to 10: {3,5,7}, {2,5,8}, or {1,5,9}. Test each by finishing the other two groups.
- {2,5,8} dies: it would leave 1, 3, 7, 9 to form two groups of 15, but 9 needs a 6 (gone) and 7 needs an 8 (gone) — impossible. The other two succeed: {1,5,9}/{3,4,8}/{2,6,7} and {3,5,7}/{1,6,8}/{2,4,9}.
- So there are 2 ways. This transfers: in equal-sum partition problems, first compute the target, then corner the extreme values — the largest elements have the fewest legal homes, so placing them first kills most of the casework.
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