Based on the the following doubt asked at facebook ……

Find the number of triangles with exactly one side odd and perimeter = 203.

The question is a little too lengthy for CAT. Yet it does allow you an opportunity to build approach & stamina when detailed lengthy work is involved.

Approach 1: You would need to know that the number of triangles with perimeter = p is …..

…..\(\left[\frac{p^2}{48}\right]\), for p being even

…..\(\left[\frac{{\left(p+3\right)}^2}{48}\right]\) for p being odd

where [x] is the nearest integer function.

Since perimeter is odd, the total number of triangles = \(\left[\frac{{\left(203+3\right)}^2}{48}\right]\)

= \(\left[\frac{206\; × \;206}{48}\right]\) = \(\left[\frac{103\; ×\; 103}{12}\right]\) = \(\left[\frac{10609}{12}\right]\) = [884.0833] = 884.

These will be triangles with (1 side odd & 2 even) or (all 3 sides odd)

If all three sides are odd, say (2a – 1), (2b – 1) and (2c – 1), then
(2a – 1) + (2b – 1) + (2c – 1) = 203
i.e. a + b + c = 103

We can consider this to be equivalent to finding triangles with perimeter 103 (e.g. triangle 34, 34, 35 will actually be 2×34-1, 2×34-1, 2×35-1 i.e. 67, 67, 69)

Since perimeter is odd (103), the total number of triangles = \(\left[\frac{{\left(103+3\right)}^2}{48}\right]\)

= \(\left[\frac{106\; ×\; 106}{48}\right]\) = \(\left[\frac{53\; × \;53}{12}\right]\) = \(\left[\frac{2809}{12}\right]\) = [234.0833] = 234.

Thus, required number of triangles with one side odd = 884 – 234 = 650

I dont like this above approach since it is heavily dependent on the formula and I do not really know if I have missed something. So my personal choice is the following, though it is more time consuming than the above.

Approach 2:

All sides have to be less than 203/2 i.e. less than 101.5 (from the triangle inequality)

Let the sides be odd, 2a and 2b. Since each side is less than 101.5, hence each of a and b has to be less than 101.5 i.e. less than 50.75

If odd side = 101, 2a + 2b = 102 i.e. a + b = 51.
Maximum value of a = 50 (& b = 1), minimum value = 1
⇒ Number of ordered solutions is 50 (ordered means a=1,b=50 is distinct from a=50,b=1).
And 25 un-ordered solutions.

If odd side = 99, 2a + 2b = 104 i.e. a + b = 52.
Maximum value of a = 50 (& b = 2), minimum value = 2
⇒ Number of ordered solutions is 49.
And 24+1 = 25 un-ordered solutions.

Let odd side = 97, 2a + 2b = 106 i.e. a + b = 53.
Maximum value of a = 50 (& b = 3), minimum value = 3
⇒ Number of ordered solutions is 48.
And 24 un-ordered solutions.

Let odd side = 95, 2a + 2b = 108 i.e. a + b = 54.
Maximum value of a = 50 (& b = 4), minimum value = 4
⇒ Number of ordered solutions is 47
And 23+1 = 24 un-ordered solutions.

And so on till a total of 51 terms (101 is the 51st odd number)

Thus, required sum = 25 + 25 + 24 + 24 + …… + 1 + 1 + 0
(This means that when odd side = 1, there will be no possible triangle. This I leave for you to check)

Answer = 2∑25 = 25 × 26 = 650.