10. (8 points) There is a sequence of numbers: $1,3,8,22,60,164,448, \cdots$ where the first number is 1, the second number is 3, and starting from the third number, each number is exactly twice the sum of the previous two numbers. In this sequence, the remainder when the 2000th number is divided by 9 is . $\qquad$
【Solution】Solution: By dividing the first few terms of the sequence by 9 and taking the remainder, we get $1, 3, 8, 4, 6, 2, 7, 0, 5, 1, 3, 8$ ... which is a repeating sequence.
$2000 \div 9$ has a remainder of 2. Since "starting from the third number, each number is exactly twice the sum of the previous two numbers," the first three numbers should be removed.

Therefore, the remainder when the 2000th number is divided by 9 is 3.