Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2020 AMC 10A Problems/Problem 10"

(→‎Solution 3 (a bit more tedious than others))
 
(24 intermediate revisions by 16 users not shown)
Line 9: Line 9:
 
==Solution 1==
 
==Solution 1==
  
−
The volume of each cube follows the pattern of <math>n^3</math> ascending, for <math>n</math> is between <math>1</math> and <math>7</math>.
+
The volume of each cube follows the pattern of <math>n^3</math>, for <math>n</math> is between <math>1</math> and <math>7</math>.
  
−
We see that the total surface area can be comprised of three parts: the sides of the cubes, the tops of the cubes, and the bottom of the <math>7\times 7\times 7</math> cube (which is just <math>7 \times 7 = 49</math>). The sides areas can be measured as the sum <math>4\sum_{n=0}^{7} n^2</math>, giving us <math>560</math>. Structurally, if we examine the tower from the top, we see that it really just forms a <math>7\times 7</math> square of area <math>49</math>. Therefore, we can say that the total surface area is <math>560 + 49 + 49 = \boxed{\textbf{(B) }658}</math>.
+
We see that the total surface area can be comprised of three parts: the sides of the cubes, the tops of the cubes, and the bottom of the <math>7\times 7\times 7</math> cube (which is just <math>7 \times 7 = 49</math>). The sides areas can be measured as the sum <math>4\sum_{n=1}^{7} n^2</math>, giving us <math>560</math>. Structurally, if we examine the tower from the top, we see that it really just forms a <math>7\times 7</math> square of area <math>49</math>. Therefore, we can say that the total surface area is <math>560 + 49 + 49 = \boxed{\textbf{(B) }658}</math>.
−
Alternatively, for the area of the tops, we could have found the sum <math>\sum_{n=0}^{6}((n+1)^{2}-n^{2})</math>, giving us <math>49</math> as well.
+
Alternatively, for the area of the tops, we could have found the sum <math>\sum_{n=2}^{7}((n)^{2}-(n-1)^{2})</math>, giving us <math>49</math> as well.
  
 
~ciceronii
 
~ciceronii
 +
 +
Note: The area on top and bottom are 49 because the largest area is 49, and the other cubes are "inscribed" in it.
 +
  
 
==Solution 2==
 
==Solution 2==
  
−
It can quickly be seen that the side lengths of the cubes are the integers from 1 to 7, inclusive.
+
It can quickly be seen that the side lengths of the cubes are the integers from 1 to 7 inclusive.
  
 
First, we will calculate the total surface area of the cubes, ignoring overlap. This value is <math>6 ( 1^2 + 2^2 + \cdots + 7^2 ) = 6\sum_{n=1}^{7} n^2 = 6 \left( \frac{7(7 + 1)(2 \cdot 7 + 1)}{6} \right) = 7 \cdot 8 \cdot 15 = 840</math>. Then, we need to subtract out the overlapped parts of the cubes. Between each consecutive pair of cubes, one of the smaller cube's faces is completely covered, along with an equal area of one of the larger cube's faces. The total area of the overlapped parts of the cubes is thus equal to <math>2\sum_{n=1}^{6} n^2 = 182</math>. Subtracting the overlapped surface area from the total surface area, we get <math>840 - 182 = \boxed{\textbf{(B) }658}</math>. ~[[User:emerald_block|emerald_block]]
 
First, we will calculate the total surface area of the cubes, ignoring overlap. This value is <math>6 ( 1^2 + 2^2 + \cdots + 7^2 ) = 6\sum_{n=1}^{7} n^2 = 6 \left( \frac{7(7 + 1)(2 \cdot 7 + 1)}{6} \right) = 7 \cdot 8 \cdot 15 = 840</math>. Then, we need to subtract out the overlapped parts of the cubes. Between each consecutive pair of cubes, one of the smaller cube's faces is completely covered, along with an equal area of one of the larger cube's faces. The total area of the overlapped parts of the cubes is thus equal to <math>2\sum_{n=1}^{6} n^2 = 182</math>. Subtracting the overlapped surface area from the total surface area, we get <math>840 - 182 = \boxed{\textbf{(B) }658}</math>. ~[[User:emerald_block|emerald_block]]
Line 25: Line 28:
 
It can be seen that the side lengths of the cubes using cube roots are all integers from <math>1</math> to <math>7</math>, inclusive.
 
It can be seen that the side lengths of the cubes using cube roots are all integers from <math>1</math> to <math>7</math>, inclusive.
  
−
Only the cubes with side length <math>1</math> and <math>7</math> have <math>5</math> faces in the surface area and the rest have <math>3</math>. Also, since the  
+
Only the cubes with side length <math>1</math> and <math>7</math> have <math>5</math> faces in the surface area and the rest have <math>4</math>. Also, since the cubes are stacked, we have to find the difference between each <math>n^2</math> and <math>(n-1)^2</math> side length as <math>n</math> ranges from <math>7</math> to <math>2</math>.
 +
 
 +
We then come up with this: <math>5(49)+13+4(36)+11+4(25)+9+4(16)+7+4(9)+5+4(4)+3+5(1)</math>.
 +
 
 +
We then add all of this and get <math>\boxed{\textbf{(B) }658}</math>.
 +
 
 +
~aryam, edited by taarunganesh
 +
 
 +
==Solution 4==
 +
Notice that the surface area of the top cube is <math>6s^2</math> and the others are <math>4s^2</math>. Then we can directly compute. The edge length for the first cube is <math>7</math> and has a surface area of <math>294</math>. The surface area of the next cube is <math>144</math>. The surface area of the next cube <math>100</math>. The surface area of the next cube is <math>64</math>. The surface area of the next cube is <math>36</math>. The surface area of the next cube is <math>16</math>. The surface area of the next cube is <math>4</math>. We then sum up <math>294+144+100+64+36+16+4</math> to get <math>\boxed{\textbf{(B) }658}</math>.
 +
~smartatmath
 +
 
 +
==Solution 5==
 +
First of all, compute the area of the sides, excluding the top and bottoms, of the cubes. The side lengths (cube root the volumes) are 1, 2, 3, 4, 5, 6, 7.
 +
Each cube's area of the sides can be calculated with <math>4(</math>area of one side<math>)</math>=<math>4(l^2)</math> so in total that is <math>4(1+4+16+...+49)</math>
 +
so
 +
<math>4(140)=560</math>
 +
the area of all the sides of the cubes is <math>560</math>.
 +
Then, calculate the bottom face of the largest cube, <math>7*7=49</math>.
 +
Now, notice that if you stack the cubes up on top of each other, and look directly down on them, the tops of the cubes showing add up to the area of the bottom cube, the 7x7. Therefore, the sum of the area of the tops of the cubes is <math>7*7=49</math>.
 +
 
 +
Now add them all up: <math>49+49+560=658.</math> Therefore, the answer is <math>\boxed{\textbf{(B) }658}</math>.
 +
 
 +
~Hermanboxcar5
  
−
cubes are stacked, we have to find the difference between each <math>n^2</math> and <math>(n-1)^2</math> side length as <math>n</math> ranges from <math>7</math> to
+
==Solution 6 (Similar to Solution 2, without notation)==
 +
Firstly, calculate the total surface area for all the cubes. Knowing that the area of each square of each cube are <math>2^2,3^2,4^2,5^2,6^2,7^2,</math> we have <math>6(1+4+9+16+25+36+49)=6(140)=840.</math> When you stack these cubes, the top square of each are subtracted by the square area of the cube above it. Meanwhile, the bottom square of each (except the 7x7x7 cube) won't be counted because of the stacking. So, we have to subtract <math>2(1+4+9+16+25+36)=2(91)=182</math> from <math>840</math>, getting <math>\boxed{\textbf{(B) }658}.</math>
  
−
<math>2</math>.
+
~Yelechi
  
−
We then come up with this: <math>5(49)+13+4(36)+11+4(25)+9+4(16)+7+4(9)+5+4(4)+3+5(1)</math>.
 
  
−
We then add all of this and get <math>\boxed{\textbf{(B) }658}</math>.
 
  
−
~aryam
+
==Video Solution 1==
 +
 
 +
Education, The Study of Everything
 +
 
 +
https://youtu.be/tIubHgoWW_M
  
−
==Video Solution==
+
==Video Solution 2==
 
https://youtu.be/JEjib74EmiY
 
https://youtu.be/JEjib74EmiY
  
 
~IceMatrix
 
~IceMatrix
 +
 +
==Video Solution 3==
 +
https://youtu.be/ZZB8KiqHbD4
 +
 +
~savannahsolver
 +
 +
== Video Solution by OmegaLearn==
 +
https://youtu.be/FDgcLW4frg8?t=3319
 +
 +
~ pi_is_3.14
  
 
==See Also==
 
==See Also==
Line 46: Line 85:
 
{{AMC10 box|year=2020|ab=A|num-b=9|num-a=11}}
 
{{AMC10 box|year=2020|ab=A|num-b=9|num-a=11}}
 
{{AMC12 box|year=2020|ab=A|num-b=6|num-a=8}}
 
{{AMC12 box|year=2020|ab=A|num-b=6|num-a=8}}
 +
 +
[[Category: Introductory Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 01:32, 3 November 2024

The following problem is from both the 2020 AMC 12A #7 and 2020 AMC 10A #10, so both problems redirect to this page.

Problem

Seven cubes, whose volumes are $1$, $8$, $27$, $64$, $125$, $216$, and $343$ cubic units, are stacked vertically to form a tower in which the volumes of the cubes decrease from bottom to top. Except for the bottom cube, the bottom face of each cube lies completely on top of the cube below it. What is the total surface area of the tower (including the bottom) in square units?

$\textbf{(A)}\ 644\qquad\textbf{(B)}\ 658\qquad\textbf{(C)}\ 664\qquad\textbf{(D)}\ 720\qquad\textbf{(E)}\ 749$

Solution 1

The volume of each cube follows the pattern of $n^3$, for $n$ is between $1$ and $7$.

We see that the total surface area can be comprised of three parts: the sides of the cubes, the tops of the cubes, and the bottom of the $7\times 7\times 7$ cube (which is just $7 \times 7 = 49$). The sides areas can be measured as the sum $4\sum_{n=1}^{7} n^2$, giving us $560$. Structurally, if we examine the tower from the top, we see that it really just forms a $7\times 7$ square of area $49$. Therefore, we can say that the total surface area is $560 + 49 + 49 = \boxed{\textbf{(B) }658}$. Alternatively, for the area of the tops, we could have found the sum $\sum_{n=2}^{7}((n)^{2}-(n-1)^{2})$, giving us $49$ as well.

~ciceronii

Note: The area on top and bottom are 49 because the largest area is 49, and the other cubes are "inscribed" in it.


Solution 2

It can quickly be seen that the side lengths of the cubes are the integers from 1 to 7 inclusive.

First, we will calculate the total surface area of the cubes, ignoring overlap. This value is $6 ( 1^2 + 2^2 + \cdots + 7^2 ) = 6\sum_{n=1}^{7} n^2 = 6 \left( \frac{7(7 + 1)(2 \cdot 7 + 1)}{6} \right) = 7 \cdot 8 \cdot 15 = 840$. Then, we need to subtract out the overlapped parts of the cubes. Between each consecutive pair of cubes, one of the smaller cube's faces is completely covered, along with an equal area of one of the larger cube's faces. The total area of the overlapped parts of the cubes is thus equal to $2\sum_{n=1}^{6} n^2 = 182$. Subtracting the overlapped surface area from the total surface area, we get $840 - 182 = \boxed{\textbf{(B) }658}$. ~emerald_block

Solution 3 (a bit more tedious than other solutions)

It can be seen that the side lengths of the cubes using cube roots are all integers from $1$ to $7$, inclusive.

Only the cubes with side length $1$ and $7$ have $5$ faces in the surface area and the rest have $4$. Also, since the cubes are stacked, we have to find the difference between each $n^2$ and $(n-1)^2$ side length as $n$ ranges from $7$ to $2$.

We then come up with this: $5(49)+13+4(36)+11+4(25)+9+4(16)+7+4(9)+5+4(4)+3+5(1)$.

We then add all of this and get $\boxed{\textbf{(B) }658}$.

~aryam, edited by taarunganesh

Solution 4

Notice that the surface area of the top cube is $6s^2$ and the others are $4s^2$. Then we can directly compute. The edge length for the first cube is $7$ and has a surface area of $294$. The surface area of the next cube is $144$. The surface area of the next cube $100$. The surface area of the next cube is $64$. The surface area of the next cube is $36$. The surface area of the next cube is $16$. The surface area of the next cube is $4$. We then sum up $294+144+100+64+36+16+4$ to get $\boxed{\textbf{(B) }658}$. ~smartatmath

Solution 5

First of all, compute the area of the sides, excluding the top and bottoms, of the cubes. The side lengths (cube root the volumes) are 1, 2, 3, 4, 5, 6, 7. Each cube's area of the sides can be calculated with $4($area of one side$)$=$4(l^2)$ so in total that is $4(1+4+16+...+49)$ so $4(140)=560$ the area of all the sides of the cubes is $560$. Then, calculate the bottom face of the largest cube, $7*7=49$. Now, notice that if you stack the cubes up on top of each other, and look directly down on them, the tops of the cubes showing add up to the area of the bottom cube, the 7x7. Therefore, the sum of the area of the tops of the cubes is $7*7=49$.

Now add them all up: $49+49+560=658.$ Therefore, the answer is $\boxed{\textbf{(B) }658}$.

~Hermanboxcar5

Solution 6 (Similar to Solution 2, without notation)

Firstly, calculate the total surface area for all the cubes. Knowing that the area of each square of each cube are $2^2,3^2,4^2,5^2,6^2,7^2,$ we have $6(1+4+9+16+25+36+49)=6(140)=840.$ When you stack these cubes, the top square of each are subtracted by the square area of the cube above it. Meanwhile, the bottom square of each (except the 7x7x7 cube) won't be counted because of the stacking. So, we have to subtract $2(1+4+9+16+25+36)=2(91)=182$ from $840$, getting $\boxed{\textbf{(B) }658}.$

~Yelechi


Video Solution 1

Education, The Study of Everything

https://youtu.be/tIubHgoWW_M

Video Solution 2

https://youtu.be/JEjib74EmiY

~IceMatrix

Video Solution 3

https://youtu.be/ZZB8KiqHbD4

~savannahsolver

Video Solution by OmegaLearn

https://youtu.be/FDgcLW4frg8?t=3319

~ pi_is_3.14

See Also

2020 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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
2020 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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 12 Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_10&oldid=231317"