AMC 10 · 2014 · #11

Grade 6 arithmeticlogic
mean-median-mode-rangeoptimizationcasework casework ↑ Prerequisites: mean-median-mode-range
📏 Medium solution 💡 3 insights
Problem
A list of whole numbers has a known mean, median, and unique most-common value. Find the largest possible entry.

Pick an answer.

(A)
24
(B)
30
(C)
31
(D)
33
(E)
35
How to solve
Strategy Extreme Principle

The total of the eleven numbers is locked at 110. To push one entry as high as possible (Tool #14, Extreme Principle), every other entry must be pushed as low as the rules allow. Tool #4 (Introduce a Variable) sorts the list as a₁ ≤ … ≤ a₁₁ so the median pins a₆ = 9 and the mode is forced into the low slots. The only real choice is how many 8s to use: more 8s raises the mode's count (letting other numbers repeat and stay small) but costs more total, so Tool #6 (Guess and Check) tests the count. Tool #3 (Eliminate Possibilities) then reads the outcome against the five answer choices — which, satisfyingly, are exactly the values the different 8-counts produce.

1STEP 1

Turn the three statistics into hard facts

The three statistics become hard facts.

sum = 110, a₆ = 9, #(8) > #(any other value)
2STEP 2

Fix the total, shrink the rest

The total is fixed, so shrink everything else.

a₁₁ = 110 - (a₁ + a₂ + … + a₁₀)
3STEP 3

Build the cheap lower half

The lower half is built as cheaply as allowed.

a₁..a₅ = 1, 1, 8, 8, 8 → #(8) = 3, sum = 26
4STEP 4

Build the cheap upper half

The upper half must respect the mode.

a₆..a₁₀ = 9, 9, 10, 10, 11, sum = 49, #(9)=#(10)=2
5STEP 5

Read off the maximum

What is left over is 35, choice (E).

a₁₁ = 110 - 75 = 35 → (E)
Answer
35
Recheck the winning list 1, 1, 8, 8, 8, 9, 9, 10, 10, 11, 35: sum = 110 so mean = 10; sorted, the 6th term is 9 so median = 9; counts are 8{:}3, 9{:}2, 10{:}2, 1{:}2, 11{:}1, 35{:}1, so 8 is the one and only mode. All three conditions hold, and 35 is the biggest listed choice, so nothing larger is on the table.
💡Key takeaway

When the total is fixed, make one number giant by squeezing every other number down to the smallest value the mean, median, and mode rules will allow.

  • Turn the three statistics into hard facts
  • Fix the total, shrink the rest
  • Build the cheap lower half
  • Build the cheap upper half
  • Read off the maximum