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2016 AMC 10A Problem 9

2016 AMC 10A Problem 9AMC 10 Step by Step

Name: Date: September 13, 2026

Sequences & Series · difficulty 2 of 5 · about 2 min · position P1-10

A triangular array of $2016$ coins has $1$ coin in the first row, $2$ coins in the second row, $3$ coins in the third row, and so on up to $N$ coins in the $N$ th row. What is the sum of the digits of $N$ ?

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

Answer

Key insight

The rows total N(N+1)/2, so N(N+1) = 4032 = 63 * 64, giving N = 63 and digit sum 9.

Solution

The total number of coins is $1 + 2 + \dots + N = \dfrac{N(N+1)}{2}$ , so
$N(N+1) = 2 \cdot 2016 = 4032.$

Since $N(N+1)$ is a product of consecutive integers, $N$ is just below $\sqrt{4032}$ . As $63^2 = 3969$ and $64^2 = 4096$ , try $N = 63$ : indeed $63 \cdot 64 = 4032$ .

The sum of the digits of $63$ is $6 + 3 = 9$ . The answer is $\boxed{\textbf{(D)}\ 9}$ .

Why this works

Triangular numbers are the standard model for "one more each row." Once the equation $N(N+1) = 4032$ is written, a quick square-root estimate pins $N$ without solving a quadratic; $N(N+1)$ sits between $N^2$ and $(N+1)^2$ , so $N = \lfloor \sqrt{4032} \rfloor$ .

The trap

Answering 63 (or its digit sum for a nearby N like 64) instead of the digit sum of N, or misestimating the square root of 4032.

Common mistakes

Answering 63 (or its digit sum for a nearby N like 64) instead of the digit sum of N, or misestimating the square root of 4032.
Using $N^2$ or $N(N-1)$ for the triangular sum.

Techniques

Bound the quantity above/below or estimate to pin it down · Set up the equation/formula and compute; no special trick needed