2018 AMC 10B Problems/Problem 8

Problem

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?

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

Solutions

Solution 1

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 $\boxed {(D) 24}$

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

Solution 4 ... Not a Solution! Just an observation.

Notice that the number of toothpicks can be found by adding all the horizontal and all the vertical toothpicks. We can see that for the case of 3 steps, there are $2(3+3+2+1)=18$ toothpicks. Thus, the equation is $2S + 2(1+2+3...+S)=180$ with $S$ being the number of steps. Solving, we get $S = 12$ , or $\boxed {(C) 12}$ . -liu4505

Solution 5 General Formula

There are $\frac{n(n+1)}{2}$ squares. Each has $4$ toothpick sides. To remove overlap, note that there are $4n$ perimeter toothpicks. $\frac{\frac{n(n+1)}{2}\cdot 4-4n}{2}$ is the number of overlapped toothpicks Add $4n$ to get the perimeter (non-overlapping). Formula is $\text{number of toothpicks} = n^2+3n$ Then you can "guess" or factor (also guessing) to get the answer $\boxed{\text{(C) }12}$ . ~bjc

Not a solution! Just an observation.

If you are trying to look for a pattern, you can see that the first column is made of 4 toothpicks. The second one is made from 2 squares: 3 toothpicks for the first square and 4 for the second. The third one is made up of 3 squares: 3 toothpicks for the first and second one, and 4 for the third one. The pattern continues like that. So for the first one, you have 0 "3 toothpick squares" and 1 "4 toothpick squares". The second is 1 to 1. The third is 2:1. And the amount of three toothpick squares increase by one every column.

The list is as follow for the number of toothpicks used... 4,4+3,4+6,4+9, and so on. 4, 7, 10, 13, 16, 19, ...

- Flutterfly

Video Solution

https://youtu.be/FbUEFq85jGE

~savannahsolver

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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