Difference between revisions of "2010 AMC 8 Problems/Problem 25"

Latest revision as of 10:12, 15 July 2024

Problem

Everyday at school, Jo climbs a flight of $6$ stairs. Jo can take the stairs $1$ , $2$ , or $3$ at a time. For example, Jo could climb $3$ , then $1$ , then $2$ . In how many ways can Jo climb the stairs?

$\textbf{(A)}\ 13 \qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24$

Solution 1

A dynamics programming approach is quick and easy. The number of ways to climb one stair is $1$ . There are $2$ ways to climb two stairs: $1$ , $1$ or $2$ . For 3 stairs, there are $4$ ways: ( $1$ , $1$ , $1$ ) ( $1$ , $2$ ) ( $2$ , $1$ ) ( $3$ )

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 $1+2+4=7$ ways to get to step 4. The pattern can then be extended: $4$ steps: $1+2+4=7$ ways. $5$ steps: $2+4+7=13$ ways. $6$ steps: $4+7+13=24$ ways.

Thus, there are $\boxed{\textbf{(E) } 24}$ ways to get to step $6.$

Video Solution by OmegaLearn

https://youtu.be/5UojVH4Cqqs?t=4560

~ pi_is_3.14

Video by MathTalks

https://youtu.be/mSCQzmfdX-g

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.

@@ Line 17: / Line 17: @@
 Thus, there are <math>\boxed{\textbf{(E) } 24}</math> ways to get to step <math>6.</math>
-==Solution 2==
-Complementary counting is also possible. Considering the six steps, Jo has to land on the last step, so there are <math>2^5=32</math> subsets (hit steps) of the other five steps. After that, subtract the number of ways to climb the steps while taking a leap of <math>4</math>, <math>5</math>, or <math>6</math>. The eight possible ways for this is (<math>4</math>, <math>1</math>, <math>1</math>), (<math>4</math>, <math>2</math>), (<math>1</math>, <math>4</math>, <math>1</math>), (<math>1</math>, <math>1</math>, <math>4</math>), (<math>2</math>, <math>4</math>), (<math>1</math>, <math>5</math>), (<math>5</math>, <math>1</math>), and (<math>6</math>)
-Altogether this makes for <math>32-8= \boxed{\textbf{(E) 24}}</math> valid ways for Jo to get to step 6.
-==Solution 3 (Recursion)==
-We can set up a recursion to solve this problem. Suppose <math>f(n)</math> represents the number of valid ways to get to the <math>n</math>th step. <math>f(0)=1</math> because there is 1 way for Jo to get to the "<math>0</math>th" step (i.e. the ground). There is <math>1</math> way to get to the first step (a <math>1</math>-step), so <math>f(1)=1</math>. There are <math>2</math> ways to get to the second step (two <math>1</math>-steps or one <math>2</math>-step). Thus, <math>f(2) = 2</math>. In general, <math>f(n) = f(n-1) + f(n-2) + f(n-3)</math>. This is because from the <math>n-3</math>th step, Jo can take a <math>3</math>-step to get to the <math>n</math>th step, from the <math>n-2</math>th step, Jo can take a <math>2</math>-step to get to the <math>n</math>th step, and from the <math>n-1</math>th step, Jo can take a <math>1</math>-step to get to the <math>n</math>th step. We now iteratively calculate values of <math>f(n)</math>.
-<math>f(0) = 1</math>
-<math>f(1) = 1</math>
-<math>f(2) = 2</math>
-<math>f(3) = f(2) + f(1) + f(0) = 2 + 1 + 1 = 4</math>
-<math>f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7</math>
-<math>f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13</math>
-<math>f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = \boxed{\textbf{(E) 24}}</math>
-~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi]
 == Video Solution by OmegaLearn ==

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