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

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.

Video Solution

https://youtu.be/XXX4_oBHuGk?t=543

~IceMatrix

https://youtu.be/jJZxzzU1bBk

~savannahsolver

See Also

2016 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2016 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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