The entire exterior of a large wooden

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The entire exterior of a large wooden

by AAPL » Tue Jul 30, 2019 5:54 am

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Manhattan Prep

The entire exterior of a large wooden cube is painted red, and then the cube is sliced into \(n^3\) smaller cubes (where \(n > 2\)). Each of the smaller cubes is identical. In terms of \(n\), how many of these smaller cubes have been painted red on at least one of their faces?

A. \(6n^2\)
B. \(6n^2 - 12n + 8\)
C. \(6n^2 - 16n + 24\)
D. \(4n^2\)
E. \(24n - 24\)

OA B
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by Scott@TargetTestPrep » Mon Aug 05, 2019 6:07 pm
AAPL wrote:Manhattan Prep

The entire exterior of a large wooden cube is painted red, and then the cube is sliced into \(n^3\) smaller cubes (where \(n > 2\)). Each of the smaller cubes is identical. In terms of \(n\), how many of these smaller cubes have been painted red on at least one of their faces?

A. \(6n^2\)
B. \(6n^2 - 12n + 8\)
C. \(6n^2 - 16n + 24\)
D. \(4n^2\)
E. \(24n - 24\)

OA B

The number of smaller cubes that have no faces painted is (n - 2)^3. Therefore, the number of smaller cubes that have at least one face painted is:

n^3 - (n - 2)^3 = n^3 - (n^3 - 6n^2 + 12n - 8) = 6n^2 - 12n + 8

Alternate Solution:

Let n = 3. Then we know we have 3^3 = 27 smaller cubes, and 26 of them (all except the innermost cube) will have at least one face that is painted red.

If we plug in 3 for n in each answer choice, we see that choice B is the only one that gives 26 as the answer: 6 * 9 - 12 * 3 + 8 = 54 - 36 + 8 = 26.

Answer: B

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by deloitte247 » Wed Aug 07, 2019 4:55 am
$$No.\ of\ cubes\ inside=\left(n-2\right)^3\ =>\ which\ have\ no\ coloured\ faces$$
Remaining cubes will have atleast 1 face coloured red
$$Remaining\ cubes=n^3-\left(n-2\right)^3\ $$
$$=n^3-\left[n^3-8-3\cdot n\cdot2\left(n-2\right)\ \right]$$
$$=n^3-\left(n^3-8-6n^2+12n\right)$$
$$=6n^2-12n+8$$

Answer = option B