104. How many and which digits will be needed to write down all integers from 1 to 100000000 inclusive?
104. Let's first consider all integers from 0 to 99999999; for those numbers that have fewer than eight digits, we will pad them with leading zeros to make them eight-digit numbers. We will thus have 100000000 eight-digit numbers, which will require, obviously, 800000000 digits for their representation. In this case, each of the 10 digits will be used an equal number of times, since they are all equally valid (zero can stand in the first place just as any other digit). Therefore, each digit will be used 80000000 times.

Now let's count how many extra zeros (i.e., zeros added at the front of numbers with fewer than eight digits) there will be. There are nine single-digit numbers (excluding zero), 99 - 9 = 90 two-digit numbers, 999 - 99 = 900 three-digit numbers, and so on. Since we added seven zeros to the left of a single-digit number, six zeros to a two-digit number, and so on, the total number of extra zeros (excluding the first number 00000000, which was written as 00000000) will be

$7 \cdot 9 + 6 \cdot 90 + 5 \cdot 900 + 4 \cdot 9000 + 3 \cdot 90000 + 2 \cdot 900000 +$

$$
+1 \cdot 9000000 = 11111103,
$$

Now, if we add 1 to the left of the first number 00000000, we will get all integers from 1 to 100000000. We see that for writing these numbers, 80000000 twos, threes, and so on up to nines are required, 80000001 ones (one extra one was added to the left of the number 00000000), and $80000000 - 11111103 = 68888897$ zeros.