Competition · AMC preparation · step 4 of 4

AMC 8 · 2015 · #23

Grade 6 logicarithmetic
Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration
sequences-arithmeticparityfraction-arithmeticlogical-deduction caseworksystematic-enumeration ↑ Prerequisites: multi-digit-arithmeticlinear-equations-one-var
📏 Long solution 💡 4 insights
Problem
Tom has 12 slips with numbers 2, 2, 2, 2.5, 2.5, 3, 3, 3, 3, 3.5, 4, 4.5 to drop into five cups A, B, C, D, E. Each cup's slip-sum must be an integer, and the five sums must be consecutive integers increasing from A to E. A 2 is already in cup E and a 3 is already in cup B. Which cup must hold the 3.5?

Pick an answer.

(A)
A
(B)
B
(C)
C
(D)
D
(E)
E

AMC 8 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

Rather than searching all placements at once, Tool #7 (Identify Subproblems) breaks the puzzle into three clean stages: (1) compute the total of all slips, (2) use that total to pin down the five required cup-sums, (3) ask where the 3.5 can legally go. Once the five target sums are known, the third stage is a textbook Tool #3 (Eliminate Possibilities) sweep — try the 3.5 in each cup and rule out any cup where the remaining slips cannot complete the required sum. The fixed slips (2 in E, 3 in B) and the parity rule for the .5 slips do most of the eliminating.

1STEP 1

Add all twelve slips

Group like slips and add: three 2s=6, two 2.5s=5, four 3s=12, plus 3.5, 4, 4.5 give a grand total of 35.

6 + 5 + 12 + 3.5 + 4 + 4.5 = 35
2STEP 2

Find each cup's target sum

Five consecutive integers averaging 35÷5=7 must be 5, 6, 7, 8, 9, so A=5, B=6, C=7, D=8, E=9.

avg = 35 ÷ 5 = 7 → A=5, B=6, C=7, D=8, E=9
3STEP 3

Test cup A

Put 3.5 in A (sum 5): leftover 5−3.5=1.5 is below the smallest slip 2, so A is out.

5 - 3.5 = 1.5 ＜ 2 = min(remaining slips) → impossible
4STEP 4

Test cup B

Put 3.5 in B (sum 6): B already holds a 3, so 3+3.5=6.5 overshoots 6, ruling B out.

3 + 3.5 = 6.5 ＞ 6 → impossible
5STEP 5

Test cup C

Put 3.5 in C (sum 7): leftover 3.5 can't be built — a lone 2.5 needs a 1 that doesn't exist, and any two slips sum to at least 4.

7 - 3.5 = 3.5; 3.5 - 2.5 = 1 ∉ slips → impossible
6STEP 6

Test cup E

Put 3.5 in E (sum 9): E already has a 2, so leftover 9−2−3.5=3.5 hits the same dead end as C, ruling E out.

9 - 2 - 3.5 = 3.5 → same dead end as cup C
7STEP 7

Confirm cup D

Only D (sum 8) remains: 8−3.5=4.5 matches the lone 4.5 slip, so 3.5 must go in cup D.

8 - 3.5 = 4.5 = available slip → (D)
Answer
D
Sanity-check the full placement implied by cup D = {3.5, 4.5}. Remaining slips: 2, 2, 2, 2.5, 2.5, 3, 3, 3, 4 (plus the fixed 2 in E and 3 in B). Cup A=5: {2, 3}. Cup B=6: already has 3, add 3 to get 6. Cup C=7: {2.5, 2.5, 2}=7. Cup E=9: already has 2, add 3 and 4 to get 9. All slips used, all sums match. Answer (D) is consistent.
💡Key takeaway

This AMC 8 problem only needs Grade 6 equation reasoning to find the cup sums, plus careful elimination — both skills you already have!

  • Add all twelve slips
  • Find each cup's target sum
  • Test cup A
  • Test cup B
  • Test cup C
  • Test cup E
  • Confirm cup D

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