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)
Line 8: Line 8:
  
 
==Solution==
 
==Solution==
 +
 +
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>.
 +
 +
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>610</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>610 + 49 + 49 = \boxed{\text{(B) }658}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 21:10, 31 January 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

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 $610$. 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 $610 + 49 + 49 = \boxed{\text{(B) }658}$.

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

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=115932"