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

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== Problem ==
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A triangular array of <math>2016</math> coins has <math>1</math> coin in the first row, <math>2</math> coins in the second row, <math>3</math> coins in the third row, and so on up to <math>N</math> coins in the <math>N</math>th row. What is the sum of the digits of <math>N</math>?
 
A triangular array of <math>2016</math> coins has <math>1</math> coin in the first row, <math>2</math> coins in the second row, <math>3</math> coins in the third row, and so on up to <math>N</math> coins in the <math>N</math>th row. What is the sum of the digits of <math>N</math>?
  
 
<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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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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==See Also==
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{{AMC10 box|year=2016|ab=A|num-b=8|num-a=10}}
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{{MAA Notice}}

Revision as of 19:46, 3 February 2016

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 $\boxed{\textbf{(D) } 9}.$

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

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