2013 AMC 8 Problems/Problem 9

Revision as of 17:52, 19 March 2018 by G1zq (talk | contribs) (Solution 2)

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 $\boxed{\textbf{(B)}}.$

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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