Difference between revisions of "2010 AMC 8 Problems/Problem 25"
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==Problem== | ==Problem== | ||
− | Everyday at school, Jo climbs a flight of <math>6</math> stairs. | + | Everyday at school, Jo climbs a flight of <math>6</math> stairs. Jo can take the stairs <math>1</math>, <math>2</math>, or <math>3</math> at a time. For example, Jo could climb <math>3</math>, then <math>1</math>, then <math>2</math>. In how many ways can Jo climb the stairs? |
<math> \textbf{(A)}\ 13 \qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24 </math> | <math> \textbf{(A)}\ 13 \qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24 </math> | ||
− | ==Solution== | + | ==Solution 1== |
− | + | A dynamics programming approach is quick and easy. The number of ways to climb one stair is <math>1</math>. There are <math>2</math> ways to climb two stairs: <math>1</math>,<math>1</math> or <math>2</math>. For 3 stairs, there are <math>4</math> ways: | |
− | |||
− | |||
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(<math>1</math>,<math>1</math>,<math>1</math>) | (<math>1</math>,<math>1</math>,<math>1</math>) | ||
(<math>1</math>,<math>2</math>) | (<math>1</math>,<math>2</math>) | ||
Line 14: | Line 11: | ||
(<math>3</math>) | (<math>3</math>) | ||
− | 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 1+2+4= | + | 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 <math>1+2+4=7</math> ways to get to step 4. The pattern can then be extended: |
− | 4 steps: 1+2+4=7 ways. | + | <math>4</math> steps: <math>1+2+4=7</math> ways. |
− | 5 steps: 2+4+7=13 ways. | + | <math>5</math> steps: <math>2+4+7=13</math> ways. |
− | 6 steps: 4+7+13=24 ways. | + | <math>6</math> steps: <math>4+7+13=24</math> ways. |
+ | |||
+ | Thus, there are <math>\boxed{\textbf{(E) } 24}</math> ways to get to step <math>6.</math> | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/5UojVH4Cqqs?t=4560 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video by MathTalks== | ||
+ | |||
+ | https://youtu.be/mSCQzmfdX-g | ||
+ | |||
+ | |||
− | |||
==See Also== | ==See Also== | ||
− | {{AMC8 box|year=2010|num-b= | + | {{AMC8 box|year=2010|num-b=24|after=Last Problem}} |
{{MAA Notice}} | {{MAA Notice}} |
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.