AMC 10 · 2024 · #10
Grade 6 arithmeticPick an answer.
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.
Turn the mean into a sum
The three unknowns must sum to nine M minus 24.8.
The mean condition fixes the decimal part of x+y+z to .2 — a one-line filter on the sum.
The mean condition is really a condition on the total of the three numbers.
▸ Why?
An average is a total shared over a count, so fixing the mean fixes the total.
▸ Why?
That total is the three numbers added together, so the condition speaks about all three at once.
Split by the range
There are only three cases.
Range = 7 has only three structural ways to happen — list them and treat each cleanly.
6.EE.B.5Identify SubproblemsNarrow the first case
The median condition pins the range of x.
Walk both feasible sums; one is killed by the median, the other narrows x to a short interval.
6.SP.B.5Make A Systematic ListFind the first solution
The first case yields one.
Push the median condition through the sorted list; in this sub-case only x = 6 survives.
6.SP.B.5Eliminate PossibilitiesFind the second solution
The second case also gives one.
Same machinery as Case I — the median equation forces y to be an integer in a narrow window.
6.SP.B.5Eliminate PossibilitiesSweep the third case
Only one median candidate works.
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 ListCount them all
That leaves 3 triples.
Three cases, three survivors — the answer count drops cleanly out of the casework.
6.SP.B.5Eliminate PossibilitiesThis AMC 12 problem only needs Grade 6 "mean / median / range" definitions plus careful case-by-case checking that you already know!