Difference between revisions of "2013 AMC 8 Problems/Problem 9"
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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> | ||
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+ | ==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: | ||
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+ | <math>1, 2, 4, 8, 16, 32, 64, 128, 256, 512, \boxed{\textbf{1024}}</math> | ||
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+ | 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> | ||
==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 18:27, 11 November 2015
Contents
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?
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 .
However, because the first term is and not
, the solution to the problem is
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:
On the 11th jump, the Hulk jumps 1024 meters > 1000 meters (1 kilometer), so our answer is the 11th jump, or
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.