Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus
During AMC 10A/12A testing, the AoPS Wiki is in read-only mode and no edits can be made.

2011 AMC 10B Problems/Problem 10: Difference between revisions

← Older editNewer edit →
Nathan wailes (talk | contribs)
2,960 edits
No edit summary
MSTang (talk | contribs)
263 edits
Line 7: Line 7:
== Solution ==
== Solution ==


The sum of the other ten elements is the same as ten <math>1</math>s. <math>10^{10}</math> is the same as <math>1</math> followed by ten <math>0</math>s. If you subtract one, it is equal to ten <math>9</math>s. Therefore if you divide the sum of the other ten elements by the largest element, it is closest 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>


== See Also==
== See Also==

Revision as of 16:08, 28 January 2014

Problem

Consider the set of numbers $\{1, 10, 10^2, 10^3, \ldots, 10^{10}\}$. 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?

$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)} 101$

Solution

The requested ratio is \[\dfrac{10^{10}}{10^9 + 10^8 + \ldots + 10 + 1}.\] Using the formula for a geometric series, we have \[10^9 + 10^8 + \ldots + 10 + 1 = \dfrac{10^{10} - 1}{10 - 1} = \dfrac{10^{10} - 1}{9},\] which is very close to $\dfrac{10^{10}}{9},$ so the ratio is very close to $\boxed{\mathrm{(B) \ } 9}.$

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. Error creating thumbnail: Unable to save thumbnail to destination

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_10&oldid=58944"