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?
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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>
 
==Solution 2==
 
We can also solve this problem by listing out how far the Hulk jumps on each jump: on the first jump, he goes 1 meter, the second jump 2 meters, and so on.  Listing out these numbers, we get:
 
 
<math>1, 2, 4, 8, 16, 32, 64, 128, 256, 512, \boxed{\textbf{1024}}</math>
 
 
On the 11th jump, the Hulk jumps 1024 meters > 1000 meters (1 kilometer), so our answer is the 11th jump, or <math>\boxed{\textbf{(C)}}.</math>
 
 
 
==Solution 3==
 
The smallest power of two that is larger or equal to 1000 is <math>2^{10}</math>, or <math>1024.</math> Thus, the answer is <math>\boxed{\textbf{(C)}}.</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}}

Latest revision as of 14:51, 26 November 2020

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)?

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

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 (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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