Difference between revisions of "2016 AMC 10A Problems/Problem 9"
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+ | == 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 1== | ||
+ | |||
+ | 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 perfect square that is close to, but less than, <math>4032</math>. Since <math>63^2 = 3969</math>, we see that <math>N = 63</math> is a likely candidate. Multiplying <math>63\cdot64</math> confirms that our assumption is correct. | ||
+ | |||
+ | == Solution 2 (Adding but somewhat more concise) == | ||
+ | |||
+ | Knowing that each row number can stand for the number of coins there are in the row, we can just add until we get <math>2016</math>. | ||
+ | Notice that <math>1 + 2 + 3 \cdots + 10 = 55.</math> Knowing this, we can say that <math>11 + 12 \cdots + 20 = 155</math> and <math>21 + \cdots +30 =255</math> and so on. This is a quick way to get to the point that N is between 60 and 70. By subtracting from the sum of the number from 1 through 70, we learn that when we subtract <math>70, 69, 68, 67, 66, 65,</math>and <math>64, N = 63.</math> Adding those two digits, we get the answer <math>\boxed{\textbf{(D) } 9}.</math> - CorgiARMY | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/XXX4_oBHuGk?t=543 | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | https://youtu.be/jJZxzzU1bBk | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC10 box|year=2016|ab=A|num-b=8|num-a=10}} | ||
+ | {{AMC12 box|year=2016|ab=A|num-b=5|num-a=7}} | ||
+ | {{MAA Notice}} |
Latest revision as of 14:34, 18 June 2022
Contents
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 1
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 perfect square that is close to, but less than, . Since , we see that is a likely candidate. Multiplying confirms that our assumption is correct.
Solution 2 (Adding but somewhat more concise)
Knowing that each row number can stand for the number of coins there are in the row, we can just add until we get . Notice that Knowing this, we can say that and and so on. This is a quick way to get to the point that N is between 60 and 70. By subtracting from the sum of the number from 1 through 70, we learn that when we subtract and Adding those two digits, we get the answer - CorgiARMY
Video Solution
https://youtu.be/XXX4_oBHuGk?t=543
~IceMatrix
~savannahsolver
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.