2016 AMC 10A Problems/Problem 9: Difference between revisions
Math101010 (talk | contribs) Created page with "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 o..." |
No edit summary |
||
| Line 1: | Line 1: | ||
== Problem == | |||
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> | ||
== Solution == | |||
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> | |||
==See Also== | |||
{{AMC10 box|year=2016|ab=A|num-b=8|num-a=10}} | |||
{{MAA Notice}} | |||
Revision as of 18:46, 3 February 2016
Problem
A triangular array of 
coins has 
coin in the first row, 
coins in the second row, 
coins in the third row, and so on up to 
coins in the th row. What is the sum of the digits of ?
Solution
We are trying to find the value of such that ![\[1+2+3\cdots+N-1+N=\frac{N(N+1)}{2}=2016.\]](http://latex.artofproblemsolving.com/4/c/2/4c218461a24de88251f68d418e63eb0c6fdae6b8.png)
Noticing that 
we have 
so our answer is 
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 | ||
These problems are copyrighted © by the Mathematical Association of America. 
