Difference between revisions of "2010 AMC 8 Problems/Problem 25"
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Thus, there are <math>\boxed{\textbf{(E) } 24}</math> ways to get to step <math>6.</math> | Thus, there are <math>\boxed{\textbf{(E) } 24}</math> ways to get to step <math>6.</math> | ||
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== Video Solution by OmegaLearn == | == Video Solution by OmegaLearn == |
Latest revision as of 11:12, 15 July 2024
Problem
Everyday at school, Jo climbs a flight of stairs. Jo can take the stairs
,
, or
at a time. For example, Jo could climb
, then
, then
. In how many ways can Jo climb the stairs?
Solution 1
A dynamics programming approach is quick and easy. The number of ways to climb one stair is . There are
ways to climb two stairs:
,
or
. For 3 stairs, there are
ways:
(
,
,
)
(
,
)
(
,
)
(
)
For four stairs, consider what step they came from to land on the fourth stair. They could have hopped straight from the 1st, done a double from #2, or used a single step from #3. The ways to get to each of these steps are ways to get to step 4. The pattern can then be extended:
steps:
ways.
steps:
ways.
steps:
ways.
Thus, there are ways to get to step
Video Solution by OmegaLearn
https://youtu.be/5UojVH4Cqqs?t=4560
~ pi_is_3.14
Video by MathTalks
See Also
2010 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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.