Dice sum 3

I have a 2-sided dice with sides labelled 1,2 1,2 , a 3-sided dice with sides labelled 1,2,3 1,2,3 , and so on, up to a 49-sided dice with sides labelled 1,2,3,...,49 1,2,3,...,49 . I simultaneously roll all 48 of them. Let di d_{i} denote the number that comes up on the dice with i i sides. The number of ways that I can roll the dice so that 3i=249di 3 \mid \displaystyle\sum_{i=2}^{49} d_{i} can be expressed in the form n!3 \dfrac{n!}{3} . What is n n ?


This is the final (and hardest) of the dice sum problems. For the previous problem, see Dice Sums 2.

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