AMC 10 · 2002 · #15

Grade 6 arithmetic
mean-median-mode-rangeoptimization extremal-construction ↑ Prerequisites: mean-median-mode-range
📏 Long solution 💡 3 insights
Problem
A collection of eight integers has its mean, median, unique mode, and range all equal to 8. Find the largest value that any single integer in the collection could be.

Pick an answer.

(A)
11
(B)
12
(C)
13
(D)
14
(E)
15
How to solve
Strategy Extreme Principle

The question asks for the biggest a single number can be, which is exactly what Tool #14 (Extreme Principle) is built for: to push one number to its maximum, force every other number down to the smallest value the rules still allow. Tool #4 (Introduce a Variable) pins the picture down — call the largest L, so the smallest is L-8 from the range. Tool #6 (Guess and Check) does the final work: with a fixed total of 64, test the top answer choices and either build a valid collection or show one is impossible. The mode clue is the brake — it stops the small numbers from all being tiny, because 8 has to appear more than anything else.

1STEP 1

Turn the four clues into number facts

Read the clues as facts: the eight numbers total 64, and largest minus smallest is 8.

sum=8 × 8=64, largest-smallest=8
2STEP 2

Aim the largest number as high as possible

To lift the largest L, make the rest minimal; the range then fixes the smallest at L-8.

largest=L, smallest=L-8
3STEP 3

Test the biggest choice, 15

If L were 15, the other seven must total 49 and each is at least 7, forcing seven 7's — mode 7, not 8.

49=7 × 7→ seven 7's→ mode=7 ≠ 8
4STEP 4

Build a collection with largest 14

With L = 14 the set 6,6,6,8,8,8,8,14 satisfies mean, median, mode and range all 8.

6,6,6,8,8,8,8,14: sum=64, median=8, mode=8, range=8
5STEP 5

Compare and conclude

15 is impossible and 14 is achieved, so the largest possible integer is 14, choice (D).

15 impossible, 14 works → (D) 14
Answer
14
The construction 6,6,6,8,8,8,8,14 is a concrete collection meeting all four conditions, so 14 is definitely achievable. And 15 was ruled out cleanly: its seven non-largest numbers would each have to be at least 7 yet total only 49, forcing them all to 7 and wrecking both the mode and the median. So 14 is both reachable and the ceiling — the answer sits right at the top choice that survives, which is a natural place for a maximization answer.
💡Key takeaway

To make one number as big as possible under a fixed total, squeeze every other number down to the smallest the rules allow — then check the biggest choice actually builds.

  • Turn the four clues into number facts
  • Aim the largest number as high as possible
  • Test the biggest choice, 15
  • Build a collection with largest 14
  • Compare and conclude