10. A sequence of numbers is arranged in a row, following this rule: the first two numbers are both 1, and starting from the third number, each number is the sum of the two preceding ones, that is: $1,2,3,5,8,13$, $21,34,55$, How many even numbers are there in the first 100 numbers of this sequence (including the 100th number)?
10.【Solution】By observing the numbers that have been written, we can see that there is an even number every two odd numbers. This pattern is not hard to explain: since the sum of two odd numbers is even, there must be an even number after two odd numbers. On the other hand, the sum of an odd number and an even number is odd, so the number after an even number is odd, and the one after that is also odd. Thus, an even number is always followed by two consecutive odd numbers, and these two odd numbers are always followed by another even number, and so on.

Therefore, even numbers appear in the third, sixth, ninth... ninety-ninth positions. The number of even numbers is equal to the number of multiples of 3 within 100, which is $99 \div 3 = 33$. Thus, there are 33 even numbers among the first 100 numbers in this sequence.

The sequence given in this problem is called the "Fibonacci sequence," also known as the "rabbit sequence."

Answer: There are 33 even numbers among the first 100 numbers in this sequence.