2018 AMC 10B Problems/Problem 8

Sara makes a staircase out of toothpicks as shown: $[asy] size(150); defaultpen(linewidth(0.8)); path h = ellipse((0.5,0),0.45,0.015), v = ellipse((0,0.5),0.015,0.45); for(int i=0;i<=2;i=i+1) { for(int j=0;j<=3-i;j=j+1) { filldraw(shift((i,j))*h,black); filldraw(shift((j,i))*v,black); } }[/asy]$ This is a 3-step staircase and uses 18 toothpicks. How many steps would be in a staircase that used 180 toothpicks?

$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 30$

Solution

A staircase with $n$ steps contains $4 + 6 + 8 + ... + 2n - 2$ steps. This can be rewritten as $(n+1)(n+2) -2$ .

So, $(n+1)(n+2) - 2 = 180$

So, $(n+1)(n+2) = 182.$

Inspection could tell us that $13 * 14 = 182$ , so the answer is $13 - 1 = \boxed {(C) 12}$

Solution 2

Layer $1$ : $4$ steps

Layer $1,2$ : $10$ steps

Layer $1,2,3$ : $18$ steps

Layer $1,2,3,4$ : $28$ steps

From inspection, we can see that with each increase in layer the difference in toothpicks between the current layer and the previous increases by 2. Using this pattern:

$4, 10, 18, 28, 40, 54, 70, 88, 108, 130, 154, 180$

From this we see that the solution is indeed $\boxed {(C) 12}$

2018 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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All AMC 10 Problems and Solutions

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2018 AMC 10B Problems/Problem 8

Solution

Solution 2

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