Let's arrange some balls

The number $N$ of ways a score of $40$ % can be achieved is equal to the coefficient of $x^{240}$ in the expression $(1 + x + x^2 + \cdots + x^{100})^6 \; = \; \left(\frac{1-x^{101}}{1-x}\right)^6 \; = \; (1 - x^{101})^6(1-x)^{-6} \; = \; \sum_{j=0}^6 (-1)^j{6 \choose j}x^{101j} \times \sum_{k=0}^\infty {k+5 \choose 5}x^k$ (for $|x| < 1$ ), and hence $N \; = \; \sum_{j=0}^2 (-1)^j {6 \choose j}{245-101j \choose 5} \; = \; 4188528351$ Thus the digit sum of $N$ is $\boxed{45}$ .