AMC 10 · 2022 · #10

Grade 6 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint

mean-median-mode-rangeconvert-to-algebralinear-equations-one-varsystematic-enumeration convert-to-algebraguess-and-checkidentify-subproblems ↑ Prerequisites: mean-median-mode-rangelinear-equations-one-var

📏 Long solution 💡 3 insights

Problem

Camila writes down five positive integers. Their unique mode is 2 more than their median, and their median is 2 more than their mean. What is the smallest possible value of the mode?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

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.

1STEP 1

Sort the five integers; the median is the middle one, M, so the mode is M + 2.

median = x₃ = M, mode = M + 2

2STEP 2

M + 2 beats the median, so both top slots equal it: x₁, x₂, M, M+2, M+2, and x₁ ＜ x₂ ＜ M keeps the mode unique.

sorted: x₁, x₂, M, M+2, M+2, 1 ≤ x₁ ＜ x₂ ＜ M

3STEP 3

Apply median = mean + 2: the sum is x₁ + x₂ + 3M + 4, and clearing the fraction gives 2M = x₁ + x₂ + 14.

2M = x₁ + x₂ + 14

4STEP 4

To shrink the mode, shrink M, hence x₁ + x₂; since 2M and 14 are both even, x₁ + x₂ must be even.

x₁ + x₂ must be even

5STEP 5

Take the smallest distinct positives with even sum, x₁ = 1 and x₂ = 3; then 2M = 18, so M = 9 (and 3 ＜ 9 ✓).

x₁ = 1, x₂ = 3 → 2M = 18 → M = 9

6STEP 6

Check {1, 3, 9, 11, 11}: mean 7, median 9, unique mode 11 — mode - median = 2 and median - mean = 2, both hold.

{1,3,9,11,11}: mean=7, median=9, mode=11

7STEP 7

Add 2 to the smallest median: the least possible mode is M + 2 = 11 — answer (D).

mode_min = 11 → (D)

Answer

Try a smaller mode and show it fails. Can mode = 9? Then M = 7 and 2M = 14, forcing x₁ + x₂ = 0 — impossible with positive integers. Can mode = 7? Then M = 5 and 2M = 10, forcing x₁ + x₂ = -4 — impossible. Can mode = 5? Then M = 3 and 2M = 6, forcing x₁ + x₂ = -8 — impossible. So choices (A), (B), (C) are all ruled out. (E) 13 would correspond to M = 11 and x₁ + x₂ = 8, which works but is not minimal. The smallest valid mode is indeed 11, matching (D).

💡Key takeaway

This AMC 10 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.

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