Difference between revisions of "2018 AMC 10B Problems/Problem 8"
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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 | ||
+ | |||
+ | === Solution 5 General Formula === | ||
+ | There are <math>\frac{n(n+1)}{2}</math> squares. Each has <math>4</math> toothpick sides. To remove overlap, note that there are <math>4n</math> perimeter toothpicks. <math>\frac{\frac{n(n+1)}{2}\cdot 4-4n}{2}</math> is the number of overlapped toothpicks | ||
+ | Add <math>4n</math> to get the perimeter (non-overlapping). Formula is <math>\text{number of toothpicks} = n^2+3n</math> Then you can "guess" or factor (also guessing) to get the answer <math>\boxed{\text{(C) }12}</math>. | ||
=== Not a solution! Just an observation. === | === Not a solution! Just an observation. === |
Revision as of 20:04, 20 November 2020
Contents
Problem
Sara makes a staircase out of toothpicks as shown:
This is a 3-step staircase and uses 18 toothpicks. How many steps would be in a staircase that used 180 toothpicks?
Solutions
Solution 1
A staircase with steps contains toothpicks. This can be rewritten as .
So,
So,
Inspection could tell us that , so the answer is
Solution 2
Layer : steps
Layer : steps
Layer : steps
Layer : 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 . Using this pattern:
From this we see that the solution is
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 and the leading coefficient is . The function is where is the layer and is the number of toothpicks.
We have to solve for when . Factor to get . The roots are and . Clearly is impossible so the answer is .
~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 toothpicks. Thus, the equation is with being the number of steps. Solving, we get , or . -liu4505
Solution 5 General Formula
There are squares. Each has toothpick sides. To remove overlap, note that there are perimeter toothpicks. is the number of overlapped toothpicks Add to get the perimeter (non-overlapping). Formula is Then you can "guess" or factor (also guessing) to get the answer .
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
~savannahsolver
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.