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

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<math>\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10</math>
 
<math>\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10</math>
  
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== Solution ==
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== Solution 1==
  
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We are trying to find the value of <math>N</math> such that <cmath>1+2+3\cdots+N-1+N=\frac{N(N+1)}{2}=2016.</cmath> Noticing that <math>\frac{63\cdot 64}{2}=2016,</math> we have <math>N=63,</math> so our answer is <math>\boxed{\textbf{(D) } 9}.</math>
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We are trying to find the value of <math>N</math> such that <cmath>1+2+3\cdots+(N-1)+N=\frac{N(N+1)}{2}=2016.</cmath> Noticing that <math>\frac{63\cdot 64}{2}=2016,</math> we have <math>N=63,</math> so our answer is <math>\boxed{\textbf{(D) } 9}.</math>
  
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Furthermore, we are attempting to solve <math>\frac{N(N+1)}{2} = 2016 \Rightarrow N(N+1) = 2016\cdot2 = 4032</math>. Approximating <math>N(N+1) \approx N^2</math>, we are looking for a square that is close, but less than <math>4032</math>. Noting that <math>64^2 = 4096</math>, we see that <math>N = 63</math> must be that answer.
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Notice that we were attempting to solve <math>\frac{N(N+1)}{2} = 2016 \Rightarrow N(N+1) = 2016\cdot2 = 4032</math>. Approximating <math>N(N+1) \approx N^2</math>, we were looking for a perfect square that is close to, but less than, <math>4032</math>. Since <math>63^2 = 3969</math>, we see that <math>N = 63</math> is a likely candidate.  Multiplying <math>63\cdot64</math> confirms that our assumption is correct.
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== Solution 2 (Adding but somewhat more concise) ==
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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>.
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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
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==Video Solution==
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https://youtu.be/XXX4_oBHuGk?t=543
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~IceMatrix
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https://youtu.be/jJZxzzU1bBk
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 +
~savannahsolver
  
 
==See Also==
 
==See Also==

Latest 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

See Also

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

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