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 perfect square that is close to, but less than, $4032$ . Since $63^2 = 3969$ , 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 $1 + 2 + 3 \cdots + 10 = 55.$ Knowing this, we can say that $11 + 12 \cdots + 20 = 155$ and $21 + \cdots +30 =255$ 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

Solution 3 (Brute force)

If you continue from solution one's conclusion that $\frac{N(N+1)}{2}=2016$ , the equation can be simplified to: ${N(N+1)}=4032$ now we can factorize 4032 into 2^6, 3^2, and 7, this means that N, and N+1 have to be made from some combination of these numbers, and we can try out values until we get that n is 63, and n+1 is 64. Adding the digits of N, we get that the answer is, $\boxed{\textbf{(D) } 9}.$ -LIUGRA001

Video Solution

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

~IceMatrix

https://youtu.be/jJZxzzU1bBk

~savannahsolver

2016 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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All AMC 10 Problems and Solutions

2016 AMC 12A (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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