Difference between revisions of "2013 AMC 8 Problems/Problem 9"
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==Problem== | ==Problem== | ||
| − | The Incredible Hulk can double the distance he jumps with each succeeding jump. If his first jump is 1 meter, the second jump is 2 meters, the third jump is 4 meters, and so on, then on which jump will he first be able to jump more than 1 kilometer? | + | The Incredible Hulk can double the distance he jumps with each succeeding jump. If his first jump is 1 meter, the second jump is 2 meters, the third jump is 4 meters, and so on, then on which jump will he first be able to jump more than 1 kilometer (1,000 meters)? |
<math>\textbf{(A)}\ 9^\text{th} \qquad \textbf{(B)}\ 10^\text{th} \qquad \textbf{(C)}\ 11^\text{th} \qquad \textbf{(D)}\ 12^\text{th} \qquad \textbf{(E)}\ 13^\text{th}</math> | <math>\textbf{(A)}\ 9^\text{th} \qquad \textbf{(B)}\ 10^\text{th} \qquad \textbf{(C)}\ 11^\text{th} \qquad \textbf{(D)}\ 12^\text{th} \qquad \textbf{(E)}\ 13^\text{th}</math> | ||
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However, because the first term is <math>2^0=1</math> and not <math>2^1=2</math>, the solution to the problem is <math>10-0+1=\boxed{\textbf{(C)}\ 11^{\text{th}}}</math> | However, because the first term is <math>2^0=1</math> and not <math>2^1=2</math>, the solution to the problem is <math>10-0+1=\boxed{\textbf{(C)}\ 11^{\text{th}}}</math> | ||
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| + | ==Video Solution== | ||
| + | https://youtu.be/Ifw2RXsqNwc ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2013|num-b=8|num-a=10}} | {{AMC8 box|year=2013|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 16:56, 5 May 2022
Contents
Problem
The Incredible Hulk can double the distance he jumps with each succeeding jump. If his first jump is 1 meter, the second jump is 2 meters, the third jump is 4 meters, and so on, then on which jump will he first be able to jump more than 1 kilometer (1,000 meters)?
Solution
This is a geometric sequence in which the common ratio is 2. To find the jump that would be over a 1000 meters, we note that 
.
However, because the first term is 
and not 
, the solution to the problem is 
Video Solution
https://youtu.be/Ifw2RXsqNwc ~savannahsolver
See Also
| 2013 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
