# 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 ?

by theworldkay 08.04.2015

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by theworldkay 08.04.2015

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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 us__ed. that is 20 of them__.

Let us arrange

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.

M / N = 24/30 = 4/5

So for minimum integer values of M and N, M = 4 and N = 5.

====

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

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:

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

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

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:

=============

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 us