AMC 8 · 2015 · #23
Grade 6 logicarithmeticPick an answer.
AMC 8 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
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.
Grouping like-value slips before adding is the Grade 5 decimal-arithmetic move that keeps the total honest.
5.NBT.B.7Identify SubproblemsFive consecutive integers averaging 35÷5=7 must be 5, 6, 7, 8, 9, so A=5, B=6, C=7, D=8, E=9.
Using the average to recover consecutive integers is the Grade 6 equation-reasoning shortcut for n+(n+1)+…+(n+4)=35.
6.EE.B.7Identify SubproblemsPut 3.5 in A (sum 5): leftover 5−3.5=1.5 is below the smallest slip 2, so A is out.
Comparing the required leftover to the smallest available slip is the fastest way to kill a candidate cup.
5.NBT.B.7Eliminate PossibilitiesPut 3.5 in B (sum 6): B already holds a 3, so 3+3.5=6.5 overshoots 6, ruling B out.
When a partial sum already overshoots, no extra slips can rescue it — adding slips can only increase the sum.
5.NBT.B.7Eliminate PossibilitiesPut 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.
To hit a half-integer total you need an odd number of .5 slips — checking which .5 combinations are available collapses the search quickly.
5.NBT.B.7Eliminate PossibilitiesPut 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.
Recognizing the same leftover sum that just failed elsewhere saves a second elimination pass.
5.NBT.B.7Eliminate PossibilitiesOnly D (sum 8) remains: 8−3.5=4.5 matches the lone 4.5 slip, so 3.5 must go in cup D.
After elimination, the only surviving option must be the answer — and a quick existence check confirms it works.
5.NBT.B.7Eliminate PossibilitiesThis AMC 8 problem only needs Grade 6 equation reasoning to find the cup sums, plus careful elimination — both skills you already have!