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The **surface area** of a **cube** is the **area** of the six squares that cover it. The **area** of one of them is a*a, or a ^{2} . Since these are all the same, you can multiply one of them by six, so the **surface area** of a **cube** is 6 times one of the sides squared.

**This problem is from AMC 10....**.

Seven cubes, whose volumes are 1,8,27,64,125,216 and 343 and 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?

Source

Competency

Difficulty

Suggested Book

*American Mathematics Competition 10 (AMC 10), {2020}, {Problem 10}*

*Surface Area of a Cube *

*7 out of 10*

** Challenges and Thrills of Pre-College Mathematics **

First hint

**The first hint can be **

** The volume of each cube follows the pattern of \(n^3\) ascending , for n is between 1 and 7.**

Second Hint

**For a cube it has three parts : the tops of the cube , the sides of the cube and the bottom of the \(7 \times 7 \times 7 \) cube (which is just \( 7 \times 7 = 49\))**

Third Hint

** The sides areas can be measured as the sum \( 4 \sum_{n=0}^{7} n^{2}\) giving 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.**

Final Step

**I think you have already got the answer for this problem but if not here is the last hint that we can use **

** that the total surface area is \( 560 + 49 + 49 = 658 \) which is the final answer for this sum.**

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