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Difference between revisions of "2013 AMC 8 Problems/Problem 9"

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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 <math>2^{10}=1024</math>.
 
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 <math>2^{10}=1024</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^{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>
  
 
==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}}

Revision as of 22:41, 27 November 2013

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?

$\textbf{(A)}\ 9^\text{th} \qquad \textbf{(B)}\ 10^\text{th} \qquad \textbf{(C)}\ 11^\text{th} \qquad \textbf{(D)}\ 12^\text{th} \qquad \textbf{(E)}\ 13^\text{th}$

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 $2^{10}=1024$.

However, because the first term is $2^0=1$ and not $2^1=2$, the solution to the problem is $10-0+1=\boxed{\textbf{(C)}\ 11^{\text{th}}}$

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

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