2013 AMC 8 Problem 9

Below is the video solution and professionally curated solution for Problem 9 of the 2013 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AMC 8 solutions, or check the answer key.

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Concepts:power of 2geometric sequence

Difficulty rating: 960

9.

The Incredible Hulk can double the distance he jumps with each succeeding jump. If his first jump is 11 meter, the second jump is 22 meters, the third jump is 44 meters, and so on, then on which jump will he first be able to jump more than 11 kilometer (1,0001,000 meters)?

9th9^\text{th}

10th10^\text{th}

11th11^\text{th}

12th12^\text{th}

13th13^\text{th}

Video solution:
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Written solution:

On the nnth jump, the Hulk jumps 2n12^{n-1} meters. We need the smallest nn for which 2n1>10002^{n-1}>1000.

Since 29=5122^9=512 and 210=10242^{10}=1024, the first jump longer than 10001000 meters has n1=10n-1=10, so n=11n=11.

Thus, C is the correct answer.

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