Difference between revisions of "2011 AMC 10B Problems/Problem 10"
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<math> \textbf{(A)}\ 1 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)} 101 </math> | <math> \textbf{(A)}\ 1 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)} 101 </math> | ||
| − | == Solution == | + | == Solution 1 == |
The requested ratio is <cmath>\dfrac{10^{10}}{10^9 + 10^8 + \ldots + 10 + 1}.</cmath> Using the formula for a geometric series, we have <cmath>10^9 + 10^8 + \ldots + 10 + 1 = \dfrac{10^{10} - 1}{10 - 1} = \dfrac{10^{10} - 1}{9},</cmath> which is very close to <math>\dfrac{10^{10}}{9},</math> so the ratio is very close to <math>\boxed{\mathrm{(B) \ } 9}.</math> | The requested ratio is <cmath>\dfrac{10^{10}}{10^9 + 10^8 + \ldots + 10 + 1}.</cmath> Using the formula for a geometric series, we have <cmath>10^9 + 10^8 + \ldots + 10 + 1 = \dfrac{10^{10} - 1}{10 - 1} = \dfrac{10^{10} - 1}{9},</cmath> which is very close to <math>\dfrac{10^{10}}{9},</math> so the ratio is very close to <math>\boxed{\mathrm{(B) \ } 9}.</math> | ||
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| + | == Solution 2 == | ||
| + | The problem asks for the value of <cmath>\dfrac{10^{10}}{10^9 + 10^8 + \ldots + 10 + 1}.</cmath> Written in base 10, we can find the value of <math>10^9 + 10^8 + \ldots + 10 + 1</math> to be <math>111111111.</math> Long division gives us the answer to be <math>\boxed{\mathrm{(B) \ } 9}.</math> | ||
== See Also== | == See Also== | ||
Revision as of 19:38, 23 September 2016
Contents
Problem
Consider the set of numbers 
. The ratio of the largest element of the set to the sum of the other ten elements of the set is closest to which integer?
Solution 1
The requested ratio is ![\[\dfrac{10^{10}}{10^9 + 10^8 + \ldots + 10 + 1}.\]](http://latex.artofproblemsolving.com/f/8/f/f8f5b2d7b0d24f1a3716487e0523cc14dafc5bcb.png)
Using the formula for a geometric series, we have ![\[10^9 + 10^8 + \ldots + 10 + 1 = \dfrac{10^{10} - 1}{10 - 1} = \dfrac{10^{10} - 1}{9},\]](http://latex.artofproblemsolving.com/5/a/0/5a0f26f9090148b1838f22b0393a29287d0cde87.png)
which is very close to 
so the ratio is very close to 
Solution 2
The problem asks for the value of Written in base 10, we can find the value of 
to be 
Long division gives us the answer to be
See Also
| 2011 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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, as part of the American Mathematics Competitions. 
