Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution)
Line 9: Line 9:
 
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>
 
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>
  
−
Furthermore, we are attempting to solve <math>\frac{N(N+1)}{2} = 2016 \Rightarrow N(N+1) = 2016</math>.
+
Furthermore, we are attempting to solve <math>\frac{N(N+1)}{2} = 2016 \Rightarrow N(N+1) = 2016\cdot = 4032</math>. Approximating <math>N(N+1) \approx N^2</math>, we are looking for a square that is close, but less than 4032. Noting that <math>64^2 = 4096</math>, we see that 63 works.
  
 
==See Also==
 
==See Also==

Revision as of 20:14, 9 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 $\boxed{\textbf{(D) } 9}.$

Furthermore, we are attempting to solve $\frac{N(N+1)}{2} = 2016 \Rightarrow N(N+1) = 2016\cdot = 4032$. Approximating $N(N+1) \approx N^2$, we are looking for a square that is close, but less than 4032. Noting that $64^2 = 4096$, we see that 63 works.

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_9&oldid=75740"