The entire surface of a solid cube with edgeof length 6 inches is painted. The cube is then cut into cubes with edge of length 1 inch. How many of the smaller cubes have paint on exactly 1 face?
I am truly clueless in even visualizing this problem.
One cube. 6 in by 6 in by 6 in. Volume is, therefore, 6 x 6 x 6, or 216 in cubed. Each face is painted: top, bottom, left, right, front, back.
They cut up the cube into smaller cubes, 1 in by 1 in by 1 in, or with a volume of 1 in cubed. Some of these new cubes have paint on them. Some don’t (they came from the middle of the old cube).
Does that help so far?
(Also do you have an answer? I got something but am not sure if it’s right.)
The answer is 96, but I still have trouble visualizing it.
Okay, the big cube has a volume of 6^3, or 216 in. The small cubes making up the big cube each have a a volume of 1 in. That means there are 216 small cubes.
What I cannot understand is that, if each of the faces are painted, and we want to count only the cubes that are touching each face, then each face has 6 rows by 6 columns, or 36 cubes. Multiply that by 6 painted faces and that gives you 216 painted cubes. I know I am thinking this wrong, but I have no other way of trying to interpret this rightly.
Read and interpret literally, exactly as written, and then you will understand the problem. Draw a cube perhaps.
To solve, note that the only cubes with 1 painted face are those that are on the outside but do not lie on an edge of the big cube.
@bodangles did you get 96?
You are on the right track…but they only want the cubes with paint on EXACTLY one face. The cubes that are on the edges or the corners will have paint on more than one face. So on each of the 6 sides, think: how many cubes are NOT edges or corners?
@BeCambridge this is incorrect - you are overcounting cubes that are on a side or corner of the big cube (more precisely, you are counting cubes that are on a corner 3 times, and cubes that are on a side but not a corner 2 times). You would have to subtract to take into account the overcounting.
However, here is a similar question that IMO, could feasibly appear on the SAT:
Q: How many cubes have * at least * one painted face?
Try solving this one afterwards. The solution I (and probably others) are thinking of is actually fairly simple.
[Here](Самое популярное порно видео смотреть бесплатно онлайн)is the question from an NYT excerpt.
Now I understand [it](Самое популярное порно видео смотреть бесплатно онлайн) a bit. Since it is cut into 216 cubes, each face must have 216/6 or 36 cubes. Now some of those cubes will indeed be making the border of not just one, but two faces - those are the outline cubes. So, there are 20 of them. Only 16 cubes touch one face, but that is only the top - what about the other 5 faces? So, I multiplied 16 by 6 and got 96.
However, I couldn’t do it without you guys. How did you think of it without all of the crap I was thinking? It took me a section’s time trying just to think this.
I did get 96! Yesss. I never try these math ones because I trust my grammar skills more.
First off, I’ve seen many problems similar to this. But it’s also just practice and/or sheer problem-solving ability (I did a lot of math contests in MS and HS). Others here that tutor the SAT may have also solved many problems like this.