2015 AMC 10A Problems/Problem 3

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 $10+12=\boxed{\textbf{(D)}\ 22}$

Solution 2

Alternatively, we can see with the $3$ -step staircase has $2[2(3)+2+1]=18$ toothpicks. Generalizing, we see that a staircase with $x$ steps has $2[2x+(x-1)+(x-2)+...+1]$ toothpicks. So, for $x=5$ steps, we have $2[2(5)+4+3+2+1]=40$ toothpicks. So our answer is $40-18=22$ or $D$ .

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2015 AMC 10A Problems/Problem 3

Contents

Problem

Solution

Solution 2

See Also