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

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<math> \textbf{(A)}\ 644\qquad\textbf{(B)}\ 658\qquad\textbf{(C)}\ 664\qquad\textbf{(D)}\ 720\qquad\textbf{(E)}\ 749 </math>
 
<math> \textbf{(A)}\ 644\qquad\textbf{(B)}\ 658\qquad\textbf{(C)}\ 664\qquad\textbf{(D)}\ 720\qquad\textbf{(E)}\ 749 </math>
  
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==Solution==
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==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> ascending, for <math>n</math> is between <math>1</math> and <math>7</math>.
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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=0}^{6}((n+1)^{2}-n^{2})</math>, giving us <math>49</math> as well.
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==Solution 2==
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It can quickly be seen that the side lengths of the cubes are the integers from 1 to 7, inclusive.
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First, we will calculate the total surface area of the cubes, ignoring overlap. This value is <math>6\sum_{n=1}^{7} n^2 = 6 ( 1^2 + 2^2 + \cdots + 7^2 ) = 6(140) = 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>.
  
 
==Video Solution==
 
==Video Solution==

Revision as of 00:09, 1 February 2020

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

Problem 10

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 $n^3$ ascending, 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 $4\sum_{n=0}^{7} n^2$, 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 $560 + 49 + 49 = \boxed{\textbf{(B) }658}$. ~ciceronii

Alternatively, for the area of the tops, we could have found the sum $\sum_{n=0}^{6}((n+1)^{2}-n^{2})$, giving us $49$ as well.

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 $6\sum_{n=1}^{7} n^2 = 6 ( 1^2 + 2^2 + \cdots + 7^2 ) = 6(140) = 840$. 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 $840 - 182 = \boxed{\textbf{(B) }658}$.

Video Solution

https://youtu.be/JEjib74EmiY

~IceMatrix

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

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