AMC 8 · 2014 · #24
Grade 6 arithmeticlogicPick an answer.
AMC 8 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
We want c₅0 and c₅1 to be as large as possible, but the budget of 252 cans is shared among all 100 customers. Tool #16 (Change Focus) flips the question: instead of "how large can c₅0, c₅1 be?", ask "how small can everyone else be?" — the cans we save by minimizing the rest are exactly the cans we can hand to the median pair. Tool #7 (Subproblems) splits the sorted list into three blocks — the 49 customers below the median, the median pair (c₅0, c₅1), and the 49 customers above — each handled separately. Tool #9 (Easier Problem) checks the logic on a tiny version (say 6 customers, 14 cans) before trusting it on 100.
Split the sorted list into three blocks: the 49 low customers, the median pair (c₅0, c₅1), and the 49 high customers, summing to 252.
Splitting at the median is a Grade 6 statistics move — the median is defined by its position in the sorted list, so the natural subproblems are the parts before, at, and after that position.
6.SP.B.5Identify SubproblemsGive each of the 49 low customers the minimum 1 can, so the low block totals 49 cans and leaves the rest for the middle.
"To make the middle bigger, push the bottom as low as it can go" — a Grade 4 multi-step word-problem reasoning step (find the smallest allowed value, then add).
4.OA.A.3Count The ComplementSubtract the low block from 252 to leave 203 cans for the median pair plus the high block (51 customers).
After locking in the low block, the remaining cans must cover the other 51 customers — a clean subtraction step from the subproblem split.
4.OA.A.3Identify SubproblemsSet all high values equal to c₅1 and share 203 among 51 as evenly as possible: 203 = 50 × 4 + 3, so 50 get 4 cans and one gets 3.
Dividing 203 by 51 with remainder is a Grade 5 long-division skill — the quotient 3 and remainder 50 tell us "50 customers get one extra can."
5.NBT.B.6Count The ComplementThe lone 3 must sit at c₅0 to keep the order, giving c₅0 = 3 and c₅1 = 4, so the median is .
Computing the median of an even-sized dataset as the average of the two middle values is the core Grade 6 statistics standard.
6.SP.B.5Count The ComplementTo make the middle of a sorted list as big as possible, squash the bottom as small as the rules allow — that frees up the most resources to lift the middle!