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1. The six faces of a cube are painted black. The cube is then cut into  smaller cubes, all the same size. How many of the smaller cubes have exactly one black face?

2. The six faces of a cube are painted black. The cube is then cut into  smaller cubes, all the same size. How many of the smaller cubes do not have any black faces?

3. The six faces of a cube are painted black. The cube is then cut into  smaller cubes, all the same size. One of the small cubes is chosen at random, and rolled. What is the probability that when it lands, the face on the top is black?

Jun 24, 2020
edited by ItzMe  Jun 24, 2020
edited by ItzMe  Jun 24, 2020

#1
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hi! can you please fix it so that i can see how many smaller cubes it's cut into?

on my screen it just says "The cube is then cut into  smaller cubes, all the same size."

Jun 24, 2020
#6
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It doesn't show me how many cubes its cut into, that's all that I see, sorry

ItzMe  Jun 24, 2020
#2
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This problem has also already been answered. please do not just give us 3 questions to answer. This is an aops question. Please tell us what you have tried and we can guide you, not give you the answer! You have to copy the aops question on to google then copy it again in the web2.0calc question page!

Jun 24, 2020
#3
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The cubes with one black face are those on the outside of the cube which are not on an edge. Omitting the outside squares on each face leaves a $3\times 3$ square of cubes which are each painted on exactly one face. Hence, there are $9$ cubes per face which are painted black on exactly one face, for a total of ....Figure out the rest, part of solution from aops for your first question.

Jun 24, 2020
#4
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Next one figure

The cubes that are not painted black on any sides are on the inside of the original cube. These cubes make up a $3\times 3\times 3$ cube,

Figure out the rest

Jun 24, 2020
#5
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Last one, part of solution, aops,

Rather than going through casework of the different kinds of cubes that could be selected and then rolled, we instead count the total possible number of faces that could come up and count the number of them that are black. There are 125 smaller cubes, each with 6 sides, so there are $125(6) = 750$ faces that could come up. Each face of the original cube has a $5\times 5$ square of black faces that might come up. Therefore, there are 25 black faces per side of the original cube, for a total of $25(6)=150$ black faces. Hence, our desired probability is figure out the rest....

Jun 24, 2020
#7
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Yes, thank you.

ItzMe  Jun 24, 2020