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 $13 \cdot 14 = 182$, 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

Solution 4

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 (HOW TO THINK CREATIVELY!!!)

https://youtu.be/8j0RvjRsjCc

~Education, the Study of Everything


Video Solution

https://youtu.be/FbUEFq85jGE

See Also

2018 AMC 10B (ProblemsAnswer KeyResources)
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

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