A white chalk cube of side of n units is painted black and then cut in

Question

A white chalk cube of side of n units is painted black and then cut into n^3 identical little cubes. What is the value of positive integer n ?

(1) There are 81 little cubes with at most 1 face painted in black.
(2) There are 44 little cubes with at least 2 face painted in black.

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shubhz2 · Answer

I think answer should be D.

For each of the statements, the required number can be expressed in terms of expression in variable n and as we have been given value of that expression, we can find out value of n in each case.

(For example: n^3-6n^2+3n = 81)

Bunuel , Please clarify

Bunuel · Answer

Similar question to practice: https://gmatclub.com/forum/a-white-chal ... 98232.html

Nidzo · Answer

A cube that has been painted and cut up into equal smaller cubes will have smaller cubes that are either painted on 3 sides, painted on 2 sides, painted on 1 side or painted on 0 sides.

(1) There are 81 little cubes with at most 1 face painted in black.

In other words, the sum of the pieces that are painted on 1 side and painted on 0 sides.

Formula for number of smaller cubes with 0 sides painted:  (n-2)^3

Formula for number of smaller cubes with 1 side painted:  6(n-2)^2

(n-2)^3 +6(n-2)^2 = 81

let (n-2) = x

x^3 + 6x^2 = 81

x^2 (x + 6) = 81

x = 3  and therfore  n = 5

SUFFICIENT

(2) There are 44 little cubes with at least 2 face painted in black.

In other words, the sum of the pieces that are painted on 2 sides and painted on 3 sides.

Formula for number of smaller cubes with 2 sides painted:  12(n-2)

Number of smaller cubes with 3 sides painted: This number will always be 8 as a cube has 8 vertices

12(n-2) + 8 = 44

12(n-2) = 36

n - 2 = 3

n = 5

SUFFICIENT

Answer D

sanjitscorps18 · Answer

(1) There are 81 little cubes with at most 1 face painted in black.
The total cubes after division is n^3
Now n^3 - cubes with more than 1 face painted black = 81
Cubes painted with more than 1 face black are Cubes on the edges and the vertices
Total number of such cubes = 4n + 8(n-2)
=> n^3 - 4n -8n + 16 = 81
=> n^3 - 12n = 65
=> n(n^2 - 12) = 65
Since 65 is only results in prime factors of 13 and 5 we get n = 5
--> Sufficient

(2) There are 44 little cubes with at least 2 face painted in black.
From above we know that cubes painted with more than 1 face black are Cubes on the edges and the vertices
Total number of such cubes = 4n + 8(n-2)
=> 4n + 8(n-2) = 44
=> 4n + 8n - 16 = 44
=> 12n = 60
=> n = 5
--> Sufficient

Option D

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A white chalk cube of side of n units is painted black and then cut in

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A white chalk cube of side of n units is painted black and then cut in

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