6. $\sum_{i=0}^{50} \sum_{j=0}^{50} \mathrm{C}_{50}^{i} \mathrm{C}_{50}^{j}$ modulo 31 is
$\qquad$ .
6.1.

Original expression $=\sum_{i=0}^{50} C_{50}^{i} \cdot \sum_{j=0}^{50} C_{50}^{j}=\left(2^{50}\right)^{2}=2^{100}$.
And $2^{5} \equiv 1(\bmod 31)$, so
Original expression $=2^{100} \equiv 1(\bmod 31)$.
Therefore, the remainder is 1.