2013 AMC 8 Problem 9

Attempt Problem 9 of the 2013 AMC 8 below, then check your answer against the video solution and professionally curated solution 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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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?
99th
1010th
1111th
1212th
1313th
Answer: C
Concepts:power of 2geometric sequence
Small hint:
The nnth jump has length 2n−12^{n-1}.
Big hint:
Compare powers of 22 with 10001000.
Video solution:
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Written solution:
On the nnth jump, the Hulk jumps 2n−12^{n-1} meters. We need the smallest nn for which 2n−1>10002^{n-1}>1000.
Since 29=5122^9=512 and 210=10242^{10}=1024, the first jump longer than 10001000 meters has n−1=10n-1=10, so n=11n=11.
Thus, C is the correct answer.
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