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The problem is as follows:

How many zeroes do we write when we write all the integers from 1 to 243 in base 3?

The given solution starts as follows:

The 1-digit numbers don't have any zeroes.

The 2-digit numbers use 2 zeroes: 10 and 20.

There are $3^2 = 9$ three-digit numbers starting with 1 and 9 starting with 2. For each leading digit, a zero appears in each digit in 9/3 = 3 of the numbers, so each has a total of 3+3 = 6 zeroes. Thus, the 3-digit numbers contain 2*6 = 12 zeroes.

I do not understand how they arrived with the fact that

"for each leading digit a zero appears in 9/3 = 3 of the numbers"

and why they are multiplying the result by 2 to get 12 at the end.

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Half of the three-digit numbers start with 1; the other half start with 2. Consider those that start with 1. The 'middle' digit (to the left of the leading 1) can be a 0 in three different cases: when the rightmost (least significant) digit is 0, 1 or 2. Similarly, the rightmost digit will be 0 for the three values (0, 1, 2) that the middle digit will take on. Thus, for the three-digit numbers starting with 1, we use a total of 6 zeroes. The same calculation applies for the three-digit numbers with a leading 2. This gives us a total of $2*6 = 12$ zeroes in the three-digit numbers.

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