Puzzle-21 (IIT-2005)

A rectangle with sides (2m-1) cm X (2n-1) cm is divided in to squares by drawing parallel lines in such a way that distance between two consecutive parallel lines is 1 cm.

The number of rectangles that can be drawn on the figure such that both the length and breadth are in odd cm is:  

(a) 4^m+n+1          (b) mn(m+1)(n+1)        (c) m²n²              (d) (m+n+1)²

Solution follows here:

Solution:

For this problem, we go in Reverse Engineering Method. We go from answer options.

We take diagrams with smaller dimensions and find out the possible number of rectangles and then check out with the answer options.

As (2m-1) and (2n-1) are odd numbers, we need to take examples such that top-most horizontal line and right-most vertical lines are tagged with odd numbers.

Simplest example:

1
0	1	2	3

In this case, 2m-1 = 3 => m=2;       2n-1 = 1 => n=1

Rectangles with odd-cm dimensions:

1cmX1cm => 3 numbers

3cmX1cm => 1 number

∴Number of rectangles with odd-cm dimensions = 4

From all the multiple choices the only possible option is (c) => m²n² = 2²1² = 4

So answer is (c) only.

But let us repeat the exercise with one more example:

1
0	1	2	3	4	5

In this case, 2m-1 = 5 => m=3;       2n-1 = 1 => n=1

Rectangles with odd-cm dimensions:

1cm X 1cm => 5 numbers

3cm X 1cm => 3 numbers

5cm X 1cm => 1 number

∴Total number of rectangles with odd-cm dimensions = 5+3+1 = 9

Even for this example, from all the multiple choices the only possible option is (c)

=> m²n² = 3²1² = 9

Answer(c)