Difference between revisions of "2016 AMC 10A Problems/Problem 9"
m (Added 12A box) |
(→Solution) |
||
| Line 8: | Line 8: | ||
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>. | ||
==See Also== | ==See Also== | ||
Revision as of 20:13, 9 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 
Furthermore, we are attempting to solve 
.
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. 
