Difference between revisions of "2016 AMC 10A Problems/Problem 9"
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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> | 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> | ||
− | 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 square that is close to, but less than, <math>4032</math>. Since | + | 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 square that is close to, but less than, <math>4032</math>. Since <math>64^2 = 4096</math>, we see that <math>N = 63</math> is a likely candidate. Multiplying <math>63\cdot64</math> confirms that our assumption is correct. |
==See Also== | ==See Also== |
Revision as of 11:35, 21 March 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
Noticing that
we have
so our answer is
Notice that we were attempting to solve . Approximating
, we were looking for a square that is close to, but less than,
. Since
, we see that
is a likely candidate. Multiplying
confirms that our assumption is correct.
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.