6. A sequence of numbers is determined by the following conditions: The first is 3, the second is 6, the third is 18, and each subsequent number is twice the sum of all preceding numbers, then the sixth number equals $\qquad$, starting from the $\qquad$th number in this sequence, each is greater than 2007.
6. Solution: The first number in this sequence is 3, the second is 6, the third is 18, the fourth is $(3+6+18) \times 2=54$, the fifth is $(3+6+18+54) \times 2=162$, the sixth is $(3+6+18+54+162) \times 2=486$.

Let the first number in this sequence be $\mathrm{a}$, then the second is $2 \mathrm{a}$, the third is $6 \mathrm{a}=2 \times 3 \times \mathrm{a}$, the fourth is $18 \mathrm{a}=2 \times 3^{2} \times \mathrm{a}$, the fifth is $54 \mathrm{a}=$, the sixth is $162 \mathrm{a}=2 \times 3^{4} \times \mathrm{a}$, the $\mathrm{n}$-th is $2 \times 3^{n-2} \times \mathrm{a}$. Since $\mathrm{a}=3$, the $\mathrm{n}$-th number can also be written as $2 \times 3^{n-1}$, meaning from the third number onwards, each number is three times the previous one. $2007 \div 486>3$, and $2007 \div 3<9$, so from the 8th number onwards, each number is greater than 2007.