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Difference between revisions of "2015 AMC 10A Problems/Problem 3"

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<math>\textbf{(A)}\ 9\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24</math>
 
<math>\textbf{(A)}\ 9\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24</math>
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<asy>
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size(150);
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defaultpen(linewidth(0.8));
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path h = ellipse((0.5,0),0.45,0.015), v = ellipse((0,0.5),0.015,0.45);
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for(int i=0;i<=2;i=i+1)
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{
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for(int j=0;j<=3-i;j=j+1)
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{
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filldraw(shift((i,j))*h,black);
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filldraw(shift((j,i))*v,black);
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}
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}</asy>
  
 
==Solution==
 
==Solution==
 
We can see that a <math>1</math>-step staircase requires <math>4</math> toothpicks and a <math>2</math>-step staircase requires <math>10</math> toothpicks. Thus, to go from a <math>1</math>-step to <math>2</math>-step staircase, <math>6</math> additional toothpicks are needed and to go from a <math>2</math>-step to <math>3</math>-step staircase, <math>8</math> additional toothpicks are needed. Applying this pattern, to go from a <math>3</math>-step to <math>4</math>-step staircase, <math>10</math> additional toothpicks are needed and to go from a <math>4</math>-step to <math>5</math>-step staircase, <math>12</math> additional toothpicks are needed. Our answer is <math>10+12=\boxed{\textbf{(D)}\ 22}</math>
 
We can see that a <math>1</math>-step staircase requires <math>4</math> toothpicks and a <math>2</math>-step staircase requires <math>10</math> toothpicks. Thus, to go from a <math>1</math>-step to <math>2</math>-step staircase, <math>6</math> additional toothpicks are needed and to go from a <math>2</math>-step to <math>3</math>-step staircase, <math>8</math> additional toothpicks are needed. Applying this pattern, to go from a <math>3</math>-step to <math>4</math>-step staircase, <math>10</math> additional toothpicks are needed and to go from a <math>4</math>-step to <math>5</math>-step staircase, <math>12</math> additional toothpicks are needed. Our answer is <math>10+12=\boxed{\textbf{(D)}\ 22}</math>
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==Solution 2==
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Alternatively, we can see with the <math>3</math>-step staircase has <math>2[2(3)+2+1]=18</math> toothpicks. Generalizing, we see that a staircase with <math>x</math> steps has <math>2[2x+(x-1)+(x-2)+...+1]</math> toothpicks. So, for <math>x=5</math> steps, we have <math>2[2(5)+4+3+2+1]=40</math> toothpicks. So our answer is <math>40-18=22</math> or <math>D</math>.
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==Solution 3==
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If one is too lazy to derive a formula for the number of picks needed for a given number of steps, one can simply see that to get to <math>4</math> steps, we add two blobs that have three picks each (the top and the right), and two more blobs that have two blocks each to form the steps. This adds <math>2\cdot3+2\cdot2=10</math> picks. Then, to get to <math>5</math> steps, we add two more edge blobs with <math>3</math> picks each and <math>3</math> more blobs that have two picks each. We add <math>2\cdot3+3\cdot2=12</math> more for a total increase of <math>10+12=\boxed{\textbf{(D)}~22}.</math>
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~Technodoggo
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==Video Solution (CREATIVE THINKING)==
 +
https://youtu.be/StaCPwJ9zSU
 +
 +
~Education, the Study of Everything
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 +
 +
 +
 +
==Video Solution==
 +
https://youtu.be/9BinymGHcUI
 +
 +
~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2015|ab=A|num-b=2|num-a=4}}
 
{{AMC10 box|year=2015|ab=A|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 02:20, 1 June 2024

Problem

Ann made a $3$-step staircase using $18$ toothpicks as shown in the figure. How many toothpicks does she need to add to complete a $5$-step staircase?

$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24$

[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]

Solution

We can see that a $1$-step staircase requires $4$ toothpicks and a $2$-step staircase requires $10$ toothpicks. Thus, to go from a $1$-step to $2$-step staircase, $6$ additional toothpicks are needed and to go from a $2$-step to $3$-step staircase, $8$ additional toothpicks are needed. Applying this pattern, to go from a $3$-step to $4$-step staircase, $10$ additional toothpicks are needed and to go from a $4$-step to $5$-step staircase, $12$ additional toothpicks are needed. Our answer is $10+12=\boxed{\textbf{(D)}\ 22}$

Solution 2

Alternatively, we can see with the $3$-step staircase has $2[2(3)+2+1]=18$ toothpicks. Generalizing, we see that a staircase with $x$ steps has $2[2x+(x-1)+(x-2)+...+1]$ toothpicks. So, for $x=5$ steps, we have $2[2(5)+4+3+2+1]=40$ toothpicks. So our answer is $40-18=22$ or $D$.

Solution 3

If one is too lazy to derive a formula for the number of picks needed for a given number of steps, one can simply see that to get to $4$ steps, we add two blobs that have three picks each (the top and the right), and two more blobs that have two blocks each to form the steps. This adds $2\cdot3+2\cdot2=10$ picks. Then, to get to $5$ steps, we add two more edge blobs with $3$ picks each and $3$ more blobs that have two picks each. We add $2\cdot3+3\cdot2=12$ more for a total increase of $10+12=\boxed{\textbf{(D)}~22}.$

~Technodoggo

Video Solution (CREATIVE THINKING)

https://youtu.be/StaCPwJ9zSU

~Education, the Study of Everything



Video Solution

https://youtu.be/9BinymGHcUI

~savannahsolver

See Also

2015 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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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