AMC 10 · 2016 · #6

Grade 6 algebra
Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search
triangular-numbersdigit-sum guess-and-check ↑ Prerequisites: triangular-numbers
📏 Medium solution 💡 2 insights
Problem
Rows grow by one item each and the whole pile has a known size. Add the digits of the last row's number.

Pick an answer.

(A)
6
(B)
7
(C)
8
(D)
9
(E)
10
How to solve
Strategy Introduce a Variable

The pile's total is the running sum 1 + 2 + … + N, so tool #4 (Introduce a Variable) names the last row N and turns the sum into one tidy formula, (N(N+1))/2. Setting that equal to 2016 gives a single equation in N. The equation N(N+1) = 4032 asks for two consecutive numbers whose product is 4032 — a perfect job for tool #6 (Guess and Check), since one well-aimed multiplication confirms the answer. Finally tool #7 (Identify Subproblems) reminds us the question has two parts: first find N, then separately add its digits — so we don't stop at N.

1STEP 1

Turn the pile into a formula

The pile has a standard formula.

1 + 2 + … + N = (N(N+1))/2 = 2016
2STEP 2

Clear the fraction

Clearing gives a product of consecutive numbers.

N(N+1) = 2 × 2016 = 4032
3STEP 3

Find the two consecutive numbers

Those two numbers are 63 and 64.

63 × 64 = 4032 → N = 63
4STEP 4

Add the digits of N

Its digits add to 9, choice (D).

6 + 3 = 9 → (D)
Answer
9
Check N = 63 directly: (63 × 64)/2 = 4032/2 = 2016, exactly the given total, so 63 is the right last row. Its digits 6 and 3 sum to 9, matching choice (D). A neighboring row would overshoot or undershoot: (62 × 63)/2 = 1953 ＜ 2016 and (64 × 65)/2 = 2080 ＞ 2016, confirming N can only be 63.
💡Key takeaway

The pile 1 + 2 + … + N equals (N(N+1))/2, so set it to 2016, find 63 × 64 = 4032, and add the digits of 63 to get 9.

  • Turn the pile into a formula
  • Clear the fraction
  • Find the two consecutive numbers
  • Add the digits of N

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