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# a solid 10cm cube is cut into 1cm cubes. these smaller cubes are then used to make the largest possible cube that looks solid from the outsi

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a solid 10cm cube is cut into 1cm cubes. these smaller cubes are then used to make the largest possible cube that looks solid from the outside, but is hollow inside. how many of the original 1cm cubes are not used to make this larger cube?

Aug 26, 2015

#2
+94619
+5

Here's my take on this one:

If we reconstructed the cube, we would have 1000 smaller cubes. Removing a "central" cube of 8 x 8 x 8 would leave us with the same size cube as before, but with 512 of the cubes removed. This would leave us with 488 cubes, which would be composed of two layers of 100 cubes on the "top" and "bottom" and 8 layers in between. And each of these layers will have (100- 64) cubes.

So   100 + 100 + 8(100- 64)   leaves 488 cubes. And the 512 we removed would total to 1000.

Aug 26, 2015

#1
+95369
+5

Well there will be 1000 cm cubes

Let the side length of the new cube be x cm

The front and back will have    $$x^2$$     cubes each

The top and bottom will have a further   x(x-2)   cubes each

and the sides will have a further   (x-2)(x-2) cubes each

So

$$\\2[x^2+x(x-2)+(x-2)^2]\le1000\\\\ 2[x^2+x^2-2x+x^2-4x+4]\le1000\\\\ x^2+x^2-2x+x^2-4x+4\le500\\\\ 3x^2-6x+4\le500\\\\ 3x^2-6x-496\le0\\\\$$

$${\mathtt{3}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{496}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{1\,497}}}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}{{\mathtt{3}}}}\\ {\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{1\,497}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}{{\mathtt{3}}}}\\ \end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{11.897\: \!028\: \!081\: \!435\: \!402\: \!3}}\\ {\mathtt{x}} = {\mathtt{13.897\: \!028\: \!081\: \!435\: \!402\: \!3}}\\ \end{array} \right\}$$

This is a concave up parabola.

It will be < 0 between the two roots.

So  the highest integer x value will be 13.

Number of cubes used will be

$$\\f(x)=2[x^2+x(x-2)+(x-2)^2]\\\\ f(13)=2[13^2+13(13-2)+(13-2)^2]\\\\ f(13)=2[169+143+121]\\\\ f(13)=2[169+143+121]=866\\\\ Left over blocks =1000-866 =\; 134 \;blocks$$

I think that method is correct

Aug 26, 2015
#2
+94619
+5

Here's my take on this one:

If we reconstructed the cube, we would have 1000 smaller cubes. Removing a "central" cube of 8 x 8 x 8 would leave us with the same size cube as before, but with 512 of the cubes removed. This would leave us with 488 cubes, which would be composed of two layers of 100 cubes on the "top" and "bottom" and 8 layers in between. And each of these layers will have (100- 64) cubes.

So   100 + 100 + 8(100- 64)   leaves 488 cubes. And the 512 we removed would total to 1000.

CPhill Aug 26, 2015
#3
+95369
0

Huh?

How big is your new cube and how many blocks are left over?

Aug 26, 2015
#4
+94619
0

The cube is exactly the same size as the original - constructed of 488 cubes  with 512 cubes removed......

Aug 26, 2015
#5
+95369
0

So isn't there enough blocks to make a bigger cube

I think that there are  :/

Aug 26, 2015
#6
+94619
0

Yep, Melody, I gotta' give you this one..I got the same answer when I thought about it......my original answer was garbage  !!!!

Sep 1, 2015