PRMO 2018 question 24 - distinct triangles

Q24. If N is the number of triangles of different shapes (i.e., not similar) whose angles are all integers (in degrees), what is N/100?

Prerequisite: Stars and bars method.
Example:
Distribute 3 apples in 3 kids.
(3+3-1)_C_(3-1) = 5_C_2 = 10
(0,0,3) => 3 variations
(1,1,1) => 1 variation
(01,2) => 6 variations
Total: 3+1+6 = 10


Solution:
By stars and bars method if we have to distribute 'n' apples in 'k' kids where a kid can receive 0 apples also, then the number of ways are:
(n+k-1)_C_(k-1).

In case of triangles we have to distribute 180 degrees in 3 angles but none of the angles can be 0.
So if the original equation is x + y + z = 180,
we rewrite it as X + Y + Z = 177 where X = x-1, Y = y-1 and Z = z-1
Now X,Y,Z can be 0 and we can apply stars and bars method.
179_C_2 = 15931

But these distributions treat (0,0,177) and (0,177,0) as different.
The original problem statement specifically asks for triangles for different shapes, so we have to treat them as same. We will have to somehow discount these duplicates.

Now, let's consider 3 cases:
1. All 3 angles are same
2. Exactly 2 angles are same => each triangle will show up 3 times
3. All 3 angles are distinct. => each triangle will show up 6 times

1. Only one case (60,60,60)
2. 
1,1,178
2,2,176
...
60,60,60
...
89,89,2

Total cases  = 89 - 1 =  88 since we have already counted (60,60,60)
Each case has 3 variations so total 88*3 = 264
All 3 angles distinct = 15931 - 264 - 1 = 15666
Distinct triangles (divide by 6) = 2611

So total distinct triangles = 2611 + 88 + 1 = 2700
2700/100 = 27 = answer.