AMC 10 · 2016 · #9

Grade 6 arithmetic
triangular-numbersdigit-sum guess-and-check ↑ Prerequisites: triangular-numbers
📏 Medium solution 💡 2 insights
Problem
Coins are stacked in a triangle: row 1 has 1 coin, row 2 has 2 coins, and so on, with row N having N coins. The whole pile holds 2016 coins. Find the number of the last row N, then add up the digits of N.

Pick an answer.

(A)
6
(B)
7
(C)
8
(D)
9
(E)
10

AMC 10 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

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

Pair the ends of 1 + 2 + … + N so each pair sums to N+1, collapsing the total into (N(N+1))/2, and set it equal to 2016.

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

Clear the fraction

Multiply both sides by 2 to clear the denominator, leaving N(N+1) = 4032, a product of two consecutive whole numbers.

N(N+1) = 2 × 2016 = 4032
3STEP 3

Find the two consecutive numbers

Two consecutive numbers near the square root of 4032 give 4032; testing, 63 × 64 hits it exactly, so N = 63.

63 × 64 = 4032 → N = 63
4STEP 4

Add the digits of N

The question wants the digit sum, not N: the digits of 63 are 6 and 3, so 6 + 3 = 9 gives (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