AMC 10 · 2024 · #15
Grade 6 arithmeticPick an answer.
AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
There are three independent conditions — range, mean-integer, median-integer — and four ways the min/max of the list can fall on {1, 7} versus {x, z}. Tool #7 (Identify Subproblems) makes the casework explicit by partitioning on "who is min, who is max". Tool #2 (Systematic List) then walks the three live cases (the fourth, min = 1, max = 7, has range 6 and is killed instantly). Within each case, the mean-integer condition pins x + y + z mod 1 to .2, the range condition pins one of x, z, and the median-integer condition pins the remaining free coordinate. Tool #3 (Eliminate Possibilities) does the final filtering — once we have a candidate, plug back and check all four conditions; tossing the ones that fail.
The six fixed numbers sum to 24.8, so an integer mean makes x + y + z ∈ {2.2, 11.2, 20.2, …}, and the range forces x ≥ 0.
The mean condition fixes the decimal part of x+y+z to .2 — a one-line filter on the sum.
6.SP.B.5Identify SubproblemsSplit by who is min and max. Since min = 1 & max = 7 gives range 6, drop it; the three live cases give z = 8, or x = 0, or z = x + 7.
Range = 7 has only three structural ways to happen — list them and treat each cleanly.
6.EE.B.5Identify SubproblemsCase I (z = 8): x + y = 3.2 puts the median at 3.2 (not integer), so drop it; x + y = 12.2 survives with 4.2 ≤ x ≤ 6.1.
Walk both feasible sums; one is killed by the median, the other narrows x to a short interval.
6.SP.B.5Make A Systematic ListPushing the median through the sorted list, only x = 6 works (median 6), giving the valid triple (6, 6.2, 8) with mean 5.
Push the median condition through the sorted list; in this sub-case only x = 6 survives.
6.SP.B.5Eliminate PossibilitiesCase II (x = 0): y + z = 2.2 fails on the median; y + z = 11.2 forces integer y = 5, so z = 6.2 — valid triple (0, 5, 6.2), mean 4.
Same machinery as Case I — the median equation forces y to be an integer in a narrow window.
6.SP.B.5Eliminate PossibilitiesCase III (z = x + 7): the median forces integer y ∈ {4, 5}. y = 4 gives x = 0.1, z = 7.1; y = 5 forces x out of [0, 1) — no solution.
Two integer choices of y, each plugged into the mean equation — one yields a legal x, the other doesn't.
6.SP.B.5Make A Systematic ListCase III's (0.1, 4, 7.1) checks out (median 4, mean 4, range 7). Across all cases there are 3 valid triples — answer (C).
Three cases, three survivors — the answer count drops cleanly out of the casework.
6.SP.B.5Eliminate PossibilitiesThis AMC 10 problem only needs Grade 6 "mean / median / range" definitions plus careful case-by-case checking that you already know!