Answer :
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 1:
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 3:
5 m - 3 p = 4 w
3 q - 5 n = 3 w
case 4:
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 case 2 : is the smallest large rectangle with minimum number of small rectangles used. that is 20 of them.
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 1:
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 3:
5 m - 3 p = 4 w
3 q - 5 n = 3 w
case 4:
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 case 2 : is the smallest large rectangle with minimum number of small rectangles used. that is 20 of them.