# What is the smallest number of 6 cm by 8 cm rectangles which can be fitted together to make a large rectangle with sides in the ratio 5?3 ?

1
by theworldkay

2015-04-09T20:18:24+05:30

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Let us arrange the small rectangles 6 cm X 8 cm all side by side.  Let M of them be  kept side by side with 6 cm facing outside.  So a rectangle of length 6 M cm and width 8 cm is formed.

Let us arrange N rows of them.  Thus a rectangle is formed with dimensions:  6 M cm X 8 N cm.    These dimensions are in the ratio  5 : 3  or  3 : 5.

case 1:   6 M :  8 N  = 5  : 3
M / N = 5/3 * 8/6 = 20/9
9 M = 20 N
As M and N are integers, the minimum values for which the above equation is satisfied, is :    N =  9  and  M  =  20.

the total number of small rectangles = M * N = 180

case 2:  6 M  :  8 N  =  3 :  5
M / N = 24/30 = 4/5
So  for minimum integer values of M and N,  M = 4 and N = 5.
the total number of small rectangles = 4 * 5 = 20.

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Instead of arranging the small rectangles all in the same manner, let us put them such that m of them are facing with 6 cm towards us, and n of them are facing us with 8 cm.  So total width of this row is : 6 m + 8n cm.

Let us arrange p rectangles with 6 cm and q rectangles with 8 cm facing the other dimension of the large rectangle.  So the other dimension of the big rectangle is:   6 p + 8 q cm.

In the previous set of solutions, above we assume n = 0 and p = 0.

6 m + 8 n :  6 p + 8 q  =  3: 5  or 5 : 3

case 3:
6 m + 8 n :  6 p + 8 q  = 3  :  5
30 m + 40 n = 18 p + 24 q
15 m + 20 n = 9 p + 12 q
3 (5 m - 3 p) = 4 ( 3 q - 5 n)        or        5 (3 m + 4 n) = 3 ( 3 p + 4 q)

Since, m and n are integers,  (let w and x be integers)

case 4:

case 5:
5 m - 3 p =  4 w
3 q - 5 n = 3 w

case 6:
3 m + 4 n = 3 x      and               3 p + 4 q = 5 x
3 ( x- m ) = 4 n
since  x - m is an integer, 3 and 4 are prime to each other,
smallest x - m:         x - m = 4,    n = 3   ,  so x >= 4.
for 3 p + 4 q = 5 x,
for x = 5,  we have p = 7 and q = 1.    smallest values of x, p and q.
we also have  p = 3, q = 4.

Hence,  m = 1,  n = 3,  p = 7, q = 1       or       m = 1, n = 3, p = 3, q = 4.

Thus finally,  we have the dimensions of the large rectangle as :  (6m+8n)  X  (6p + 8q)
= 30 cm X 50 cm        or    30 cm X 50 cm.

in that case : number of small rectangles: (m+n) * (p+q) = 4 * 8 = 32
or  4 * 7 = 28
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Thus  the Case 2 :  in the first discussed pattern of uniform arrangment of small rectangles in M columns and N  rows,  is the smallest large rectangle with minimum number of small rectangles used. that is 20 of them.