# 2016 AMC 10A Problems/Problem 9

## Problem

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 $N$? $\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$

## Solution 1

We are trying to find the value of $N$ such that $$1+2+3\cdots+(N-1)+N=\frac{N(N+1)}{2}=2016.$$ Noticing that $\frac{63\cdot 64}{2}=2016,$ we have $N=63,$ so our answer is $\boxed{\textbf{(D) } 9}.$

Notice that we were attempting to solve $\frac{N(N+1)}{2} = 2016 \Rightarrow N(N+1) = 2016\cdot2 = 4032$. Approximating $N(N+1) \approx N^2$, we were looking for a square that is close to, but less than, $4032$. Since $64^2 = 4096$, we see that $N = 63$ is a likely candidate. Multiplying $63\cdot64$ confirms that our assumption is correct.

## Solution 2 (Adding but somewhat more concise)

Knowing that each row number can stand for the number of coins there are in the row, we can just add until we get 2016. Notice that 1 + 2 + 3 ... + 10 = 55. Knowing this, we can say that 11 + 12 ... + 20 = 155 and 21 + ... +30 =255 and so on. This is a quick way to get to the point that N is between 60 and 70. By subtracting from the sum of the number from 1 through 70, we learn that when we subtract 70, 69, 68, 67, 66, 65,and 64, N = 63. Adding those two digits, we get the answer $\boxed{\textbf{(D) } 9}.$ - CorgiARMY

## Video Solution

~IceMatrix

~savannahsolver

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