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 1: / Line 1: @@
 ==Problem==
-Everyday at school, Jo climbs a flight of <math>6</math> stairs. Joe can take the stairs <math>1,2</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?
+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>
-==Solution==
+==Solution 1==
-We will systematically enumerate all of the possibilities. We will denote a specific climb by using a sequence of <math>1s</math>, <math>2s</math>, and <math>3s</math> in which each sequence adds to <math>6</math>.
+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:
+(<math>1</math>,<math>1</math>,<math>1</math>)
+(<math>1</math>,<math>2</math>)
+(<math>2</math>,<math>1</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 <math>1+2+4=7</math> ways to get to step 4. The pattern can then be extended:
+<math>4</math> steps: <math>1+2+4=7</math> ways.
+<math>5</math> steps: <math>2+4+7=13</math> 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
-Since there is only one way to create a sequence which contains all <math>1s</math>, all <math>2s</math>, or all <math>3s</math>, there are three possible sequences which only contain one number. If we attempt to create sequences which contain one <math>2</math> and the rest <math>1s</math>, the sequence will contain two <math>2s</math> and four <math>1s</math>. We can place the <math>2</math> in either the first, second, third, fourth, or fifth position, giving a total of five possibilities. If we attempt to create sequences which contain one <math>3</math> and the rest <math>1s</math>, the sequence will contain one <math>3</math> and three <math>1s</math>. We can place the <math>3</math> in either the first, second, third, or fourth position, giving a total of four possibilities. For sequences which contain exactly two <math>2s</math> and the rest <math>1s</math>, the sequence will contain two <math>2s</math> and two <math>1s</math>. The two <math>2s</math> could be next to each other, separated by one <math>1</math> in between, or separated by two <math>1s</math> in between. We can place the two <math>2s</math> next to each other in three ways, separated by one <math>1</math> in two ways, and separated by two <math>1s</math> in only one way. This gives us a total of six ways to create a sequence which contains two <math>2s</math> and two <math>1s</math>.
-Note that we cannot have a sequence of only <math>2s</math> and <math>3s</math> since the sum will either be <math>5</math> or greater than <math>6</math>. We now only need to consider the case where we use all three numbers in the sequence. Since all three numbers add to <math>6</math>, the number of permutations of the three numbers is <math>3!=6</math>.
-Adding up the number of sequences above, we get: <math>3+5+4+6+6=24</math>. Thus, answer choice <math>\boxed{\textbf{(E)}\ 24}</math> is correct.
 ==See Also==
-{{AMC8 box|year=2010|num-b=25|after=Last Problem}}
+{{AMC8 box|year=2010|num-b=24|after=Last Problem}}
+{{MAA Notice}}

Page

Toolbox

Search