2013 AMC 8 Problems/Problem 9

Problem

The Inedible Bulk 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 $10-0+1=\boxed{\textbf{(C)}\ 11^{\text{th}}}$

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:

$1, 2, 4, 8, 16, 32, 64, 128, 256, 512, \boxed{\textbf{1024}}$

On the 11th jump, the Hulk jumps 1024 meters > 1000 meters (1 kilometer), so our answer is the 11th jump, or $\boxed{\textbf{(C)}}.$

Solution 3

The smallest power of two that is larger or equal to 1000 is $2^{10}$ , or $1024.$ Thus, the answer is $\boxed{\textbf{(B)}}.$

2013 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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All AJHSME/AMC 8 Problems and Solutions

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2013 AMC 8 Problems/Problem 9

Contents

Problem

Solution

Solution 2

Solution 3

See Also