2015 AMC 8 Problem 23

Attempt Problem 23 of the 2015 AMC 8 below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2015 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

23.

Tom has twelve slips of paper which he wants to put into five cups labeled A,A, B,B, C,C, D,D, E.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 AA to E.E. The numbers on the papers are 2,2, 2,2, 2,2, 2.5,2.5, 2.5,2.5, 3,3, 3,3, 3,3, 3,3, 3.5,3.5, 4,4, 4.5.4.5. If a slip with 22 goes into cup EE and a slip with 33 goes into cup B,B, then the slip with 3.53.5 must go into what cup?
A A
B B
C C
D D
E E
Answer: D
Concepts:logical deductioncasework
Small hint:
The five cup sums must be consecutive integers with total 35.35.
Big hint:
After filling cup B,B, test where the 3.53.5 slip can still leave possible sums.
Video solution:
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Written solution:
The sum of all the slips is 35,35, so the five consecutive integer cup sums must average 7.7. Therefore cups A,A, B,B, C,C, D,D, and EE must have sums 5,5, 6,6, 7,7, 8,8, and 9,9, respectively.
Cup BB already contains a 33 and must sum to 6,6, so it must contain another 3.3. Cup EE already contains a 2,2, so the other slips in EE must sum to 7.7.
The 3.53.5 slip cannot go in A,A, because cup AA would need another 1.5.1.5. It cannot go in B,B, which is already full. It cannot go in CC or E,E, because either would then need another 3.5,3.5, and no remaining slips can make that total. Cup DD works, for example with 3.5+4.5=8.3.5+4.5=8.
Thus, D is the correct answer.
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