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Difference between revisions of "2018 AMC 10B Problems/Problem 8"
m (→‎Solution 2)
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<math> 4, 10, 18, 28, 40, 54, 70, 88, 108, 130, 154, 180 </math>
 
<math> 4, 10, 18, 28, 40, 54, 70, 88, 108, 130, 154, 180 </math>
  
−
From this we see that the solution is indeed <math>\boxed {(C) 12}</math>
+
From this we see that the solution is <math>\boxed {(C) 12}</math>
  
 
By: Soccer_JAMS
 
By: Soccer_JAMS

Revision as of 14:26, 16 January 2019

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 $4 + 6 + 8 + ... + 2n + 2$ toothpicks. 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 $\boxed {(C) 12}$

By: Soccer_JAMS

Solution 3

We can find a function that gives us the number of toothpicks for every layer. Using finite difference, we know that the degree must be $2$ and the leading coefficient is $1$. The function is $f(n)=n^2+3n$ where $n$ is the layer and $f(n)$ is the number of toothpicks.


We have to solve for $n$ when $n^2+3n=180\Rightarrow n^2+3n-180=0$. Factor to get $(n-12)(n+15)$. The roots are $12$ and $-15$. Clearly $-15$ is impossible so the answer is $\boxed {(C) 12}$.

~Zeric Hang

See Also

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