, 2018

For a positive integer to be divisible by 9, the sum of its digits must be divisible by 9.

In this problem, we want to count the number of six-digit positive integers containing 2018 and divisible by 9.

Thus, we must find the remaining two digits, which together with 2018, form a six-digit positive integer that is divisible by 9.

The digits 2018 have a sum of $2+0+1+8=11$.

Let the remaining two digits be $a$ and $b$ so that the six-digit positive integer is $ab2018$ or $ba2018$ or $a2018b$ or $b2018a$ or $2018ab$ or $2018ba$.

When the digits $a$ and $b$ are added to 11, the sum must be divisible by 9.

That is, the sum $a+b+11$ must be divisible by 9.

The smallest that each of $a$ and $b$ can be is 0, and so the smallest that the sum $a+b+11$ can be is $0+0+11=11$.

The largest that each of $a$ and $b$ can be is 9, and so the largest that the sum $a+b+11$ can be is $9+9+11=29$.

The only integers between 11 and 29 that are divisible by 9 are 18 and 27.

Therefore, either $a+b=18-11=7$ or $a+b=27-11=16$.

If $a+b=7$, then the digits $a$ and $b$ are 0 and 7 or 1 and 6 or 2 and 5 or 3 and 4, in some order.

If $a$ and $b$ are 1 and 6, then the possible six-digit integers are $162\,018, 612\,018, 120\,186, 620\,181,$

$201\,816,$ and $201\,861$.

In this case, there are 6 possible six-digit positive integers.

Similarly, if $a$ and $b$ are 2 and 5, then there are 6 possible six-digit integers.

Likewise, if $a$ and $b$ are 3 and 4, then there are 6 possible six-digit integers.

If $a$ and $b$ are 0 and 7, then the possible six-digit integers are $702\,018, 720\,180, 201\,870$, and $201\,807$, since the integer cannot begin with the digit 0.

In this case, there are 4 possible six-digit positive integers.

Therefore, for the case in which the sum of $a$ and $b$ is 7, there are $6+6+6+4=22$ possible six-digit integers.

Finally, we consider the case for which the sum of the digits $a$ and $b$ is 16.

If $a+b=16$, then the digits $a$ and $b$ are 7 and 9, or 8 and 8.

If $a$ and $b$ are 7 and 9, then there are again 6 possible six-digit integers ($792\,018, 972\,018,$

$720\,1869, 920\, 187, 201\,879,$ and $201\,897$).

If $a$ and $b$ are 8 and 8, then there are 3 possible six-digit integers: $882\,018, 820\,188,$ and $201\,888$.

Therefore, for the case in which the sum of $a$ and $b$ is 16, there are $6+3=9$ possible six-digit integers, and so there are $22+9=31$ six-digit positive integers in total.

We note that all of these 31 six-digit positive integers are different from one another, and that they are the only six-digit positive integers satisfying the given conditions.

Therefore, there are 31 six-digit positive integers that are divisible by 9 and that contain the digits 2018 together and in this order.