Difference between revisions of "2011 AMC 10B Problems/Problem 10"
(→Solution 2) |
Erics son07 (talk | contribs) (→Solution 2) |
||
Line 12: | Line 12: | ||
== Solution 2 == | == 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>1111111111.</math> Long division gives us the answer to be <math>\boxed{\mathrm{(B) \ } 9}.</math> | 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>1111111111.</math> Long division gives us the answer to be <math>\boxed{\mathrm{(B) \ } 9}.</math> | ||
+ | |||
+ | Alternate finish: multiply the denominator by 9 and notice that it is 1 less than <math>10^{10}</math>. So the answer is very very close to <math>\boxed{\mathrm{(B) } 9}</math>. | ||
+ | |||
+ | ~JH. L | ||
== See Also== | == See Also== |
Revision as of 04:40, 25 June 2022
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 Using the formula for a geometric series, we have 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
Alternate finish: multiply the denominator by 9 and notice that it is 1 less than . So the answer is very very close to .
~JH. L
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.