To begin with basics, a cube/cuboid has 6 surfaces
(top, bottom, left, right, front and back), 8 corners (highlighted in oval) and 12 edges (marked with arrows ). Ref Fig1

**Minimum & Maximum pieces from ‘n’ cuts**

A cube/cuboid is a 3 dimensional figure and when cut
from different dimensions, the resultant number of pieces are obtained on
multiplying the number of pieces on each dimension. Hence to minimize the
number of pieces from certain number of cuts made on the cube, they should all
be assigned on the same dimension (refer Fig 2).

As in Fig 2, a cube has been cut 3 times and as all
cuts have been made on the same dimension, the number of pieces are (3 + 1) =
4. Similarly for any

**‘n’ number of cuts, the minimum number of pieces**is obtained by assigning them on the same dimension (i.e. all across length/breadth/height) and result in**(n + 1) pieces**
In Fig 2, instead of 3 if we put any other number of
cuts the result shall be one more number of pieces.

When trying to maximize the number of pieces with
‘n’ number of cuts, the cuts should be made as equally as possible across all
dimensions. Eg. for 3 cuts – 1 each through length, breadth and height (Fig 3);
for 4 cuts – 1 through 2 dimensions and 2 through 1 (Fig 4); for 5 cuts – 1
through 1 dimension and 2 through 2 (Fig 5).

In figure 3, 1 cut on each dimension implies (1+1)
pieces on each dimension. The total number of pieces is the product of pieces
on each dimension – (2 x 2 x 2) = 8.

Having distributed 3 cuts equally across length,
breadth and height, the fourth cut is made on one of the dimensions leading to
2 pieces on two of the dimensions and one extra piece i.e. 3 on the dimension
which got an additional cut.

In case of 5 cuts, 2 cuts on 2 of the dimensions
leave 3 pieces on each of them and 1 cut on the third dimension leads to 2
pieces on that dimension. Hence total number of pieces is –

(3 x 3 x 2) = 18

#1. If a cube is cut 10 times, what is the maximum
and minimum pieces that can result from it?

Here minimum pieces
shall be (10 + 1) = 11 and maximum will be when the cuts are distributed as
equally as possible i.e. 3, 3, 4 across the three dimensions leading to –

(3 + 1) x (3 + 1) x (4
+ 1) = 80 pieces.

#2. What is the maximum
number of small pieces that a cube can be cut into with17 cuts?

Here the cuts will be
distributed as equally as possible across all dimensions in the order 5, 6, 6
which shall result in (5 + 1) x (6+ 1) x (6 + 1) = 294 pieces.

**Minimum and Maximum cuts for ‘n’ pieces**

In case the number of pieces into which a cube or
cuboid is cut into are given, then the maximum cuts that may be required to get
them shall be one less than the number of pieces given. This theory is just the
reverse of what we discussed above.

As 3 cuts gave 4 pieces on same dimension in figure
2, to get 4 pieces maximum cuts will be required when they are on the same
dimension (length/breadth/height), hence (4 – 1).

In order to find the minimum cuts for any number of
pieces the following steps need to be followed:

i) Find the nearest cube number ≥ the given number
of pieces (let’s say the cube number is ‘A’)

ii) Identify the number of cuts required to get A
(by distributing the cube root number of pieces across all dimensions and
reducing it by 1 to get the number of cuts per dimension)

iii) If A=number of pieces then the answer will be
in (ii), else the number of cuts will be reduced one by one to find the
resultant pieces

#3. What is the maximum and minimum number of times
that a cube must be cut to get 85 pieces?

Here maximum cuts will be (85 – 1) = 84

To find minimum cuts –

i) Nearest cube number ≥ 85 is 125

ii) For 125 pieces minimum cuts will be 4 + 4 + 4 =
12

iii) If cuts are reduced on every dimension one by
one then,

Cut = 4 + 4 + 3 = 11; Pieces = 5 x 5 x 4 = 100 (≥
85)

Cuts = 4 + 3 + 3 = 10; Pieces = 5 x 4 x 4 = 80 (<
85)

Hence minimum cuts required to get 85 pieces are 11
(which shall also be answer for any number of pieces from 81 – 100

**Colour on Surfaces**

When a cube/cuboid is coloured on its surfaces and then
cut into smaller cubes, the number of coloured surfaces of the small cubes vary
according to its positions.

Let us take for instance the cuboid in fig 6 –

The cuboid has been cut twice through length, thrice
through breadth and four times through height and has been cut into (3 x 4 x 5)
= 60 pieces. If the surface of the cuboid is coloured then,

__Number of small cubes coloured on three surfaces:__

*Only the small cubes on conrers have 3 surfaces exposed, which shall get coloured; hence there are*

**8 such cubes**.

__Number of small cubes coloured on two surfaces:__
The small cubes on the edge (barring the corner
ones) are the ones which have been coloured on 2 surfaces each. As there are 12
edges, there are exactly 4 edges on the length which have 3 cubes on it (lets
call it ‘L’), 4 edges on the breadth, with 4 cubes on it (say ‘B’) and 4 edges
on height with 5 cubes on it (will refer as ‘H’).

Each edge has two cubes on the ends which are corner
ones and have been coloured on 3 surfaces; hence while adding the small cubes
on edges the two corners need to be deducted.

The number of cubes with exactly 2 surfaces coloured
shall be –

4 (L-2) + 4 (B-2) + 4 (H-2) =

**4[(L-2) + (B-2) + (H-2)]**
In this case it will be 4[(3-2) + (4-2) + (5-2)] =
24

__Number of small cubes coloured on one surface:__
Small cubes which are on the surface of the cuboid
and not on edges are coloured on exactly 1 surface.

Total cubes on a surface can be calculated by
multiplying any two dimensions in pair eg. (LxB) or (BxH) or (HxL); there being
exactly two opposite surfaces for each pair. However, in order to avoid the
edges the top-bottom rows and left-right columns need to be deducted from the calculation
as follows -

2[(L-2) x (B-2)] + 2[(B-2) x (H-2)] + 2[(H-2) x
(L-2)]

=

**2[{(L-2) x (B-2)} + {(B-2) x (H-2)} + {(H-2) x (L-2)}]**
=2[{(3-2) x (4-2)} + {(4-2) x (5-2)} + {(5-2) x
(3-2)}]

= 22

__Number of small cubes without any colour:__
Small cubes which didn’t get coloured are the ones
inside the ones on the surfaces. In order to get the small cubes on surfaces
out of the way, length, breadth and height all need to be reduced by 2 to
remove the top, bottom, left, right, front and back surfaces. This can be
calculated as follows –

**(L-2) x (B-2) x (H-2)**

= (3-2) x (4-2) x (5-2)

= 6

This can also be reconfirmed by subtracting the sum
of small cubes coloured on 1, 2 and 3 surfaces from the total number of small
cubes i.e. 60 – (8 + 24 + 22) = 6.

## No comments:

## Post a Comment