Difference between revisions of "2016 AMC 10A Problems/Problem 9"

Revision as of 14:34, 18 June 2022

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

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
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 (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.

@@ Line 13: / Line 13: @@
 == 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.
+Knowing that each row number can stand for the number of coins there are in the row, we can just add until we get <math>2016</math>.
-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 <math>\boxed{\textbf{(D) } 9}.</math> - CorgiARMY
+Notice that <math>1 + 2 + 3 \cdots + 10 = 55.</math> Knowing this, we can say that <math>11 + 12 \cdots + 20 = 155</math> and <math>21 + \cdots +30 =255</math> 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 <math>70, 69, 68, 67, 66, 65,</math>and <math>64, N = 63.</math> Adding those two digits, we get the answer <math>\boxed{\textbf{(D) } 9}.</math> - CorgiARMY
 ==Video Solution==

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