C1. The positive integer $N$ has six digits in increasing order. For example, 124689 is such a number.
However, unlike 124689 , three of the digits of $N$ are 3,4 and 5 , and $N$ is a multiple of 6 . How many possible six-digit integers $N$ are there?
Solution
The digits of $N$ increase, so the known, consecutive digits 3,4 and 5 must be adjacent and in increasing order.
Since $N$ has six digits, and they increase, the final digit must be at least 5 .
Since $N$ is a multiple of 6 , it is even, so the final digit is even and must be 6 or 8 . This means that 3 (as the start of 345 ) must be one of the first three digits of $N$.
This gives the following possible values of $N$ :
$$
345678, \quad 134568, \quad 134578, \quad 234568, \quad 234578, \quad 123456 \text { and } 123458
$$

Since $N$ is a multiple of 3 , the sum of the digits of $N$ must be a multiple of 3 .
The possibilities $134578,234568,234578$ and 123458 have digit sums 28, 28, 29 and 23 respectively so cannot be values for $N$.
But $345678,134568,123456$ have digit sums 33,27 and 21 respectively.
Therefore these are the three possible six-digit integers $N$.
[An alternative interpretation of the question allows digits in the solutions to repeat, since such a number would have digits that are in increasing order, even though the digits themselves do not always increase. This interpretation yields 19 solutions.]