2011 AMC 10B Problem 10

Below is the professionally curated solution for Problem 10 of the 2011 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AMC 10B solutions, or check the answer key.

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Concepts:geometric sequenceestimation

Difficulty rating: 1280

10.

Consider the set of numbers {1,10,102,103,,1010}.\{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?

11

99

1010

1111

101 101

Solution:

The largest number is 1010.10^{10}. The rest of the number have a sum of S=i=0910i. S=\sum_{i=0}^9 10^i. Then, 10S=i=0910i+1,10S = \sum_{i=0}^9 10^{i+1}, making 9S=10101.9S = 10^{10}-1. This means that 10109S\dfrac{10^{10}}{9S} is close to one, so the ratio between 101010^{10} and the sum is close to 9.9.

Thus, the correct answer is B .

Problem 10 in Other Years