🇺🇸 AMC 8 ⇄ switch contest
2015 AMC 8

Problem 23

Problem 23 · 2015 AMC 8 Hard
Counting & Probability caseworkconstraint-satisfaction

Tom has twelve slips of paper which he wants to put into five cups labeled A, B, C, D, E. He wants the sum of the numbers on the slips in each cup to be an integer. Furthermore, he wants the five integers to be consecutive and increasing from A to E. The numbers on the papers are 2, 2, 2, 2.5, 2.5, 3, 3, 3, 3, 3.5, 4, and 4.5. If a slip with 2 goes into cup E and a slip with 3 goes into cup B, then the slip with 3.5 must go into what cup?

Show answer
Answer: D — Cup D.
Show hints
Hint 1 of 3
Before worrying about the 3.5, lock down the five cup totals. The slips add to 35, and the totals are five consecutive integers — five consecutive integers summing to 35 have middle (=average) 35/5 = 7, so they're forced.
Still stuck? Show hint 2 →
Hint 2 of 3
Once the totals are A=5, B=6, C=7, D=8, E=9, use the two given placements to remove slips, then test the 3.5 in each cup: it works only if its leftover partners can hit the cup's remaining total.
Still stuck? Show hint 3 →
Hint 3 of 3
Cup B has a 3 and needs 6, so its partner is another 3; cup E has a 2 and needs 7 more. Now ask, for each cup, 'total minus 3.5 — can the remaining slips make that?'
Show solution
Approach: pin down cup totals, then place 3.5 by elimination
  1. Sum of slips: 2+2+2+2.5+2.5+3+3+3+3+3.5+4+4.5 = 35. Five consecutive integers summing to 35 must be 5, 6, 7, 8, 9, so A=5, B=6, C=7, D=8, E=9.
  2. B has a 3 and needs 6 total ⇒ the other slip in B is another 3.
  3. After B = {3, 3} and the 2 in E, the slips still to place are {2, 2, 2.5, 2.5, 3, 3, 3.5, 4, 4.5}. Try the 3.5 in each remaining cup and see what its partner(s) would have to total.
  4. A (need 5): 5 − 3.5 = 1.5 — no slip equals 1.5 and no combo of leftovers sums to 1.5. ✗
  5. C (need 7): 7 − 3.5 = 3.5 from leftovers — no such combo without using the only 3.5. ✗
  6. E (already has 2, need 7 more): 7 − 3.5 = 3.5 — same issue. ✗
  7. D (need 8): 8 − 3.5 = 4.5 — pair with the 4.5 slip. ✓ The remaining slips then fill the others: A = {2.5, 2.5} = 5, C = {3, 4} = 7, and E takes {2, 2, 3} on top of its 2 for 9.
  8. So 3.5 goes in cup D.
  9. Why this transfers: when totals are 'consecutive integers' (or otherwise constrained), find them first from the grand sum — that collapses an open-ended placement puzzle into a small, finite check.
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