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

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==Solution==
 
==Solution==
This is a geometric sequence in which the common ratio is 2. To find the jump that would be over a 1000 meters, you have to use the power of 2 to find that the
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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}=24</math>.
                                                                            <math>2^{10}</math> = 1024.  
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However, because <math>2^0</math> and not <math>2^1</math> is the first term, the solution to the problem = <math>\boxed{\textbf{(C)}\ 11th}</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)}\ 11th}</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 19:47, 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 $2^{10}=24$.

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

See Also

2013 AMC 8 (ProblemsAnswer KeyResources)
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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