10. (12 points) There are now two animals, mice and rabbits, which grow in the following ways: every month, the number of mice doubles from the previous month, and the number of rabbits becomes the sum of the numbers from the previous two months (the second month and the first month have the same number). For example: in the first month, there are 2 mice and 2 rabbits, in the second month, there are 4 mice and 2 rabbits, in the third month, there are 8 mice and 4 rabbits. Now we know that in the 7th month, there is one more rabbit than mice. Then, in the first month, the minimum number of rabbits is $\qquad$.
【Answer】Solution: According to the problem:
To minimize the number of rabbits, the number of mice is at least 1;
The number of mice doubles in the second month, quadruples in the third month, and so on, until it becomes 64 times in the seventh month, which is 64.
The number of rabbits is 1 in the first month, 1 in the second month, 2 in the third month, 3 in the fourth month, 5 in the fifth month, 8 in the sixth month, and 13 in the seventh month.

We get $(64+1) \div 15=5$.
Therefore, the answer is: 5