Given the following equation

$S_{1}+S_{2}+S_{3}+S_{4}+S_{5}+S_{6} = 30$ such that $S_{i} \in \mathbb{N} \cup \{0\} \space \forall i \in [6] $

How many integer solutions exist, subject to the constraint:

$\forall i \space S_{i}=i \pmod 3$

Work so far:

Since $S_{1}=10, S_2=2, S_3=3,S_4=4,S_5=5,S_6=6$ is a solution, there is at least one solution. Also, without the constraint, the number of integral solutions is ${30 + 6-1 \choose 6-1} = {35 \choose 5}$ therefore we know that the number $n$ of integral solutions under the constraint is $1\leq n \leq {35 \choose 5}$


Write $S_i = 3a_i+i\pmod 3$, so we have$$3a_1+1+3a_2+2+3a_3+0+3a_4+1+3a_5+2+3a_6=30$$

where $a_i\geq 0$, so $$a_1+a_2+...+a_6 = 8$$

and thus we have $${8+5\choose 5}$$ solutions

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