Amount of visible cubes

iren.perch

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Jan 5, 2020
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On a wooden table is a cube with rib 6 which exists of 6^3 identical cubes with rib 1. How many of these small cubes are visible?
I thought:
6^3 - 4^3, but that's foult...
Can someone help me?
Thanks in advance.
 
On a wooden table is a cube with rib 6 which exists of 6^3 identical cubes with rib 1. How many of these small cubes are visible?
I thought:
6^3 - 4^3, but that's foult...
Can someone help me?
Thanks in advance.
The corner and the edge cubes are being counted multiple times by your method.
 
On a wooden table is a cube with rib 6 which exists of 6^3 identical cubes with rib 1. How many of these small cubes are visible?
I thought:
6^3 - 4^3, but that's foult...
Can someone help me?
Thanks in advance.
I think you mean this:

On a wooden table is a cube with side 6, which consists of 6^3 identical cubes with side 1. How many of the small cubes are visible?​

You have correctly counted the number of cubes that are exposed to the outside from any direction, ignoring that fact that the cube is sitting on a table, which blocks the view of the bottom.

You can use the same method with a small modification. Which cubes are not visible in the given situation? It is not only those on the inside ...
 
Thank you both!
@SubhotoshKhan
How can I bypass that problem?
@Dr.Peterson
If you mean I should do minus another 25, than that answer is foult too...
I have the choice between:
(A) 132 (B) 136 (C) 140 (D) 156 (E) 180
 
Think more about what Dr. Peterson said here:

"Which cubes are not visible in the given situation? It is not only those on the inside ... "
 
Thank you both!
@SubhotoshKhan
How can I bypass that problem?
@Dr.Peterson
If you mean I should do minus another 25, than that answer is foult too...
I have the choice between:
(A) 132 (B) 136 (C) 140 (D) 156 (E) 180
First, SK was wrong. Your method avoids double-counting, and is almost correct.

But in saying 25, I think you have the right idea but are just being hasty. Think carefully about how many cubes on the bottom can't be seen at all. One of their answers is indeed correct.

By the way, by "foult" do you mean "faulty"? I would just say "wrong".
 
Thank you both!
@SubhotoshKhan
How can I bypass that problem?
@Dr.Peterson
If you mean I should do minus another 25, than that answer is foult too...
I have the choice between:
(A) 132 (B) 136 (C) 140 (D) 156 (E) 180
Just looking from top (plan view) - how many cube due you see - 6 * 6 (=36)

Imagine those were colored black (remember to color the edge cubes and corner cubes - properly).

Now look from front (elevation view) - how many new cubes do you see? (= 5 *6) color those Yellow.

Now look from back (elevation view) - how many new cubes do you see? (= 5 *6) color those Yellow.

Continue with looking from left-side and right-side ,,,,,,,
 
First, SK was wrong. Your method avoids double-counting, and is almost correct.

But in saying 25, I think you have the right idea but are just being hasty. Think carefully about how many cubes on the bottom can't be seen at all. One of their answers is indeed correct.

By the way, by "foult" do you mean "faulty"? I would just say "wrong".
Sorry for my English and -4*4?
 
First, SK was wrong. Your method avoids double-counting, and is almost correct.

But in saying 25, I think you have the right idea but are just being hasty. Think carefully about how many cubes on the bottom can't be seen at all. One of their answers is indeed correct.

By the way, by "foult" do you mean "faulty"? I would just say "wrong".
Yes ... I was wrong - I mis-read the question. Jomo .... shut up - I am going to the corner
 
Another way to get this answer is to see that the invisible cubes form a cuboid 4 by 4 by 5 (removing a layer on each side and the top, which are visible, but including the bottom). So the number is 6*6*6 - 4*4*5 = 216 - 80 = 136.
 
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