AMC 10 · 2014 · #18

Grade 6 arithmeticlogic
Archetype Pair-Sum Invariant └ Sub-archetype Universe Total Minus Known
mean-median-mode-rangeoptimizationcasework casework ↑ Prerequisites: mean-median-mode-range
📏 Medium solution 💡 3 insights
Problem
A list of 11 positive integers has mean 10, median 9, and a unique mode of 8. Find the largest value any single integer in the list can be.

Pick an answer.

(A)
24
(B)
30
(C)
31
(D)
33
(E)
35

AMC 10 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

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

Mean 10 across 11 numbers forces the total 110; the median pins a₆ = 9, and since 8 is smaller, every 8 sits in slots a₁ to a₅.

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

Fix the total, shrink the rest

The pot is stuck, so the biggest entry is 110 minus the other ten — squeeze those ten to the smallest the rules permit.

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

Build the cheap lower half

Three 8s in the low slots, the other two filled with 1, the smallest positive integer: 1, 1, 8, 8, 8 sums to 26.

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

Build the cheap upper half

From a₆ up nothing may reach three copies, so the cheapest legal run is 9, 9, 10, 10, 11 — sum 49.

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

Read off the maximum

The first ten total 26 + 49 = 75, so the last and largest entry is 110 - 75 = 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

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