Difference between revisions of "2018 AMC 10B Problems/Problem 8"

(Solution 4 ... Not a Solution! Just an observation.)
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So, <math>(n+1)(n+2) = 182.</math>  
 
So, <math>(n+1)(n+2) = 182.</math>  
  
Inspection could tell us that <math>13 * 14 = 182</math>, so the answer is <math>13 - 1 = \boxed {(C) 12}</math>
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Inspection could tell us that <math>13 \cdot 14 = 182</math>, so the answer is <math>13 - 1 = \boxed {(C) 12}</math>
  
 
=== Solution 2 ===
 
=== Solution 2 ===
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By: Soccer_JAMS
 
By: Soccer_JAMS
  
=== Solution 3 ===
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===Solution 3 (10th Grade Math)===
 +
 
 +
Notice that we have the points \( (1,4) \), \( (2,10) \), \( (3,18) \). Then, we calculate the first difference to get 6 and 8 respectively, and calculate the second difference to get 2. This is a quadratic expression in the form \( ax^2 + bx + c \).
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Because we have a table of elements, \( 2a = \) the second difference, so \( 2a = 2 \Rightarrow a = 1 \). The quadratic is now \( y = x^2 + bx + c \). We plug in two points to get \( 4 = b + c \) and \( 10 = 4 + 2b + c \Rightarrow 6 = 2b + c \). Solving through elimination gives us \( b = 2 \), and \( c = 2 \).
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 +
We now have \( y = x^2 + 2x + 2 \). We put \( y = 180 \), and see that \( 180 = x^2 + 2x + 2 \), and \( 0 = x^2 + 2x - 178 \). We use the quadratic formula to get \( -2 \pm \sqrt{716}/2 \). \( \sqrt{716} \) is greater than \( 26^2 \) but less that \( 27^2 \), so we say \( \sqrt{716} = 26 \). We then get 12 or -14.
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We cannot have a negative number of steps, so our answer is <math>\boxed {(C) 12}</math>.
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~Pinotation
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 +
=== Solution 4 ===
 
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 <math>2</math> and the leading coefficient is <math>1</math>. The function is <math>f(n)=n^2+3n</math> where <math>n</math> is the layer and <math>f(n)</math> is the number of toothpicks.
 
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 <math>2</math> and the leading coefficient is <math>1</math>. The function is <math>f(n)=n^2+3n</math> where <math>n</math> is the layer and <math>f(n)</math> is the number of toothpicks.
  
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~Zeric Hang
 
~Zeric Hang
  
=== Not a Solution! Just an observation. ===
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=== Solution 5 ===
 
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 <math>2(3+3+2+1)=18</math> toothpicks. Thus, the equation is <math>2S + 2(1+2+3...+S)=180</math> with <math>S</math> being the number of steps. Solving, we get <math>S = 12</math>, or <math>\boxed {(C) 12}</math>.
 
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 <math>2(3+3+2+1)=18</math> toothpicks. Thus, the equation is <math>2S + 2(1+2+3...+S)=180</math> with <math>S</math> being the number of steps. Solving, we get <math>S = 12</math>, or <math>\boxed {(C) 12}</math>.
 
-liu4505
 
-liu4505
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The list is as follow for the number of toothpicks used...
 
The list is as follow for the number of toothpicks used...
4,4+3,4+6,4+9, and so on.
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<math>4</math>,<math>4+3</math>,<math>4+6</math>,<math>4+9</math>, and so on.
 
4, 7, 10, 13, 16, 19, ...
 
4, 7, 10, 13, 16, 19, ...
  
 
- Flutterfly
 
- Flutterfly
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 +
==Video Solution (HOW TO THINK CREATIVELY!!!)==
 +
https://youtu.be/8j0RvjRsjCc
 +
 +
~Education, the Study of Everything
 +
 +
  
 
=== Video Solution ===
 
=== Video Solution ===
 
https://youtu.be/FbUEFq85jGE
 
https://youtu.be/FbUEFq85jGE
 
~savannahsolver
 
  
 
== See Also ==
 
== See Also ==
 
{{AMC10 box|year=2018|ab=B|num-b=7|num-a=9}}
 
{{AMC10 box|year=2018|ab=B|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:57, 2 September 2025

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 (10th Grade Math)

Notice that we have the points \( (1,4) \), \( (2,10) \), \( (3,18) \). Then, we calculate the first difference to get 6 and 8 respectively, and calculate the second difference to get 2. This is a quadratic expression in the form \( ax^2 + bx + c \).

Because we have a table of elements, \( 2a = \) the second difference, so \( 2a = 2 \Rightarrow a = 1 \). The quadratic is now \( y = x^2 + bx + c \). We plug in two points to get \( 4 = b + c \) and \( 10 = 4 + 2b + c \Rightarrow 6 = 2b + c \). Solving through elimination gives us \( b = 2 \), and \( c = 2 \).

We now have \( y = x^2 + 2x + 2 \). We put \( y = 180 \), and see that \( 180 = x^2 + 2x + 2 \), and \( 0 = x^2 + 2x - 178 \). We use the quadratic formula to get \( -2 \pm \sqrt{716}/2 \). \( \sqrt{716} \) is greater than \( 26^2 \) but less that \( 27^2 \), so we say \( \sqrt{716} = 26 \). We then get 12 or -14.

We cannot have a negative number of steps, so our answer is $\boxed {(C) 12}$.

~Pinotation

Solution 4

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 5

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

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