12. Write the consecutive odd numbers from 1 to 103 as a single large number: \( A = \) 13579111315171921….9799101103.
Then the number \( \mathrm{a} \) has \_\_\_\_ digits, and the remainder when \( \mathrm{a} \) is divided by 9 is \_\_\_\_ _\.
12、 $101 ; 4$
Solution: There are 5 one-digit odd numbers, 45 two-digit odd numbers, and adding two three-digit odd numbers, so $\mathrm{a}$ is a $5+2 \times 45+3 \times 2=101$ (digit) number.

Starting from 1, the remainders when consecutive odd numbers are divided by 9 are $1,3,5,7,0,2,4,6,8$, $1,3,5,7,0,2,4,6,8, \cdots$. Starting from 1, each cycle consists of 9 numbers: $1,3,5,7,0,2,4,6,8$. Since $(1+3+5+7+0+2+4+6+8)$ is divisible by 9, from 1 to 89, it is exactly 5 cycles. Therefore, the remainder when this 101-digit number $\mathrm{a}$ is divided by 9 is the remainder when $1+3+5+7+0+2+4$ is divided by 9, which is 4.

Solution 2: The remainder when a natural number is divided by 9 is the same as the remainder when the sum of its digits is divided by 9. Using this property, in $a=13579111315171921 \cdots \cdots .9799101103$, the sum of the digits of 13579, when divided by 9, gives a remainder of 7, while the sum of the digits of $111315171921 \cdots \cdots .9799$, when divided by 9, gives a remainder of 0, and the sum of the digits of 101103, when divided by 9, gives a remainder of 6. Therefore, the remainder when $a$ is divided by 9 is the remainder when $(7+6)$ is divided by 9, which is 4.