AMC 10 · 2008 · #12

Grade 6 algebra
Archetype Arithmetic Expression Decomposition └ Sub-archetype Linear Rate Accumulation
mean-median-mode-rangedifference-of-squaresperfect-squares convert-to-algebra ↑ Prerequisites: mean-median-mode-range
📏 Short solution 💡 2 insights
Problem
For every count, the average of that many terms equals the count itself. Find one particular term far down the list.

Pick an answer.

(A)
2008
(B)
4015
(C)
4016
(D)
4030056
(E)
4032064
How to solve
Strategy Introduce a Variable

The average of a group equals its total divided by how many things are in it. So I give the running total a name, S_n, the sum of the first n terms. The average rule turns into a clean formula for S_n. Then a single term is just the jump from one running total to the next, which splits the task into a small subproblem I can compute.

1STEP 1

Turn the average into a sum

Each average becomes a clean running total.

S_n = (mean) × n = n × n = n²
2STEP 2

One term is the jump between totals

One term is the jump between two of those totals.

a₂₀₀₈ = S₂₀₀₈ - S₂₀₀₇ = 2008² - 2007²
3STEP 3

Factor the difference of squares

That difference of squares factors to 4015, choice (E).

2008² - 2007² = (2008-2007)(2008+2007) = 1 × 4015 = 4015
Answer
4015
Check the pattern on small cases. S₁ = 1, so term 1 is 1. S₂ = 4, so term 2 is 4-1=3. S₃ = 9, so term 3 is 9-4=5. The terms are 1, 3, 5, 7, …, the odd numbers, and the nth odd number is 2n-1. For n=2008 that gives 2(2008)-1 = 4015, matching (B).
💡Key takeaway

If the average of the first n terms is always n, the running total is n², so each term is the gap between two neighbouring squares.

  • Turn the average into a sum
  • One term is the jump between totals
  • Factor the difference of squares

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