Question

How many 6 digit numbers have exactly 3 different digits?

Answer 1

First imagine 6 numbers arranged in rising sequence.
The median of 6 numbers being 10 implies that there are 3 numbers bigger than 10 and 3 numbers smaller than 10. Consider the 3rd and 4th number.
To have the median of 10, these have to be smaller or bigger than 10 by the same amount. So the possible pairs can be 1 and 19, 2 and 18 and so on all the way to 9 and 11. Furthermore, since they had to be distinct, 1 and 2 cannot be the 3rd number, so the smallest 3rd number is 3, and the largest is 9.
How does this influence the mean?
If the 3rd number is 3, the 1st and 2nd numbers must be 1 and 2. What’s more, the 4th number being 17, the 5th number must be at least 18. For the mean to be 13, the 6th (largest) number would be:

13*6–1–2–3–17–18=37

If the 3rd number is 9, the 1st and 2nd numbers can be 1 and 2 (this allows for the largest possible 6th number for the mean of 13). The 4th number is 11 and the 5th number can be 12. So the 6th number would be:
13*6–1–2–9–11–12=43

And this is the solution. The largest possible number in the sequence is 43. The sequence of natural numbers is 1,2,9,11,12,43