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Difference between revisions of "2015 AMC 10A Problems/Problem 3"

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<math>\textbf{(A)}\ 9\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24</math>
 
<math>\textbf{(A)}\ 9\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24</math>
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path h = ellipse((0.5,0),0.45,0.015), v = ellipse((0,0.5),0.015,0.45);
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for(int i=0;i<=2;i=i+1)
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{
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for(int j=0;j<=3-i;j=j+1)
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filldraw(shift((i,j))*h,black);
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filldraw(shift((j,i))*v,black);
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}</asy>
  
 
==Solution==
 
==Solution==

Revision as of 17:16, 17 July 2015

Problem

Ann made a $3$-step staircase using $18$ toothpicks as shown in the figure. How many toothpicks does she need to add to complete a $5$-step staircase?

$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24$

[asy] size(150); defaultpen(linewidth(0.8)); path h = ellipse((0.5,0),0.45,0.015), v = ellipse((0,0.5),0.015,0.45); for(int i=0;i<=2;i=i+1) { for(int j=0;j<=3-i;j=j+1) { filldraw(shift((i,j))*h,black); filldraw(shift((j,i))*v,black); } }[/asy]

Solution

We can see that a $1$-step staircase requires $4$ toothpicks and a $2$-step staircase requires $10$ toothpicks. Thus, to go from a $1$-step to $2$-step staircase, $6$ additional toothpicks are needed and to go from a $2$-step to $3$-step staircase, $8$ additional toothpicks are needed. Applying this pattern, to go from a $3$-step to $4$-step staircase, $10$ additional toothpicks are needed and to go from a $4$-step to $5$-step staircase, $12$ additional toothpicks are needed. Our answer is $10+12=\boxed{\textbf{(D)}\ 22}$

See Also

2015 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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 AMC 10 Problems and Solutions

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