AMC 10 · 2022 · #7
Grade 6 arithmeticPick an answer.
The problem packages three statistics (mode, median, mean) at once. Tool #7 (Subproblems) — break it into 'what does the sorted list look like?', 'what equation does the mean give?', and 'how small can the unknowns be?'. Tool #13 (Algebra) — once the shape is fixed, one equation in one unknown M tells the whole story. Tool #6 (Guess and Check) — try the smallest legal x₁, x₂ first; if parity fails, bump up. We avoid heavier inequality machinery — small directional guesses on positive integers find the floor cleanly.
Place the median and mode
The mode takes the top two slots.
In a sorted list of five, the median is just the third number — Grade 6 statistics.
6.SP.B.5Identify SubproblemsWrite them sorted
Write all five sorted.
The mode must show up twice in the top slots; the bottom three must all be different so no other value ties the mode.
6.SP.A.3Identify SubproblemsUse the mean condition
The mean fixes the total.
Turning the mean condition into one clean linear equation in M, x₁, x₂.
6.EE.B.7Convert To AlgebraSpot the parity condition
The bottom two must add to an even number.
Parity of both sides forces x₁ + x₂ to be even.
Because the total has to be a whole multiple of the count, the two free entries must add to an even number.
▸ Why?
An average is a total shared over a count, so the total is forced to be that count times the mean.
▸ Why?
Odd and even totals cannot both work, so the parity of the free part is pinned down.
Make it as small as possible
Take the bottom two as small as possible.
Smallest legal x₁, x₂ with even sum makes M smallest — guess-and-check on positive integers.
5.NBT.B.5Guess And CheckCheck a real example
Verify with an actual set of five.
All three conditions hold — the construction is valid.
6.SP.B.5Guess And CheckRead the minimum
The smallest mode is 11.
Add 2 to the minimum median to read off the minimum mode.
4.NBT.B.4Identify SubproblemsThis AMC 12 problem only needs Grade 6 mean-median-mode and one-step equations you already know — sort the list as x₁, x₂, M, M+2, M+2, set up 2M = x₁ + x₂ + 14, pick the smallest distinct positive x₁, x₂ with even sum (1 and 3), and the mode lands at M + 2 = 11.