It's well known that the addition of internal angles for a polygon is 180(n-2) and for a regular one each angle will be 180(n-2)/n now we need the addition of all the polygons to be 360. Sum \sum_{i=1}^{k} 180(n_i-2)/n_i = 360 => \sum_{i=1}^{k} 1-2/n_i = 2 => \sum_{i=1}^{k} 1/n_i = (k-2)/2 If there were more than 6 polygons that would mean that there is one with angle lower to 60, but this isn't true, therefore we just need to check the cases from k=3 untill k=6 which I won't do here. (It didn't let me post the sums because i'm new just add the $ and it will show)