We only need to look at prime number, since if a number isn’t divisible by p, it won’t be divisible by any multiple of p.

We’ll start at the bottom and work our way up. Since the number always ends with 9, it will never be divisible by 2.

For divisibility by 3, the sum of the digits would need to be divisible by 3, but since the digits 6 and 9 are divisible by 3, the leading 1 means the number as a whole will never be.

Since the number ends with 9 it will never be divisible by 5.

Next let’s look at 7. If we multiply a number of the form 16..9 by 3 we will always get a number of the form 50..7. if we then subtract 7 we will have a number that is clearly just the product of 2s and 5s. Since multiplying by 3 and subtracting 7 would have preserved any divisibility by 7, this proves that 16..9 is never a multiple of 7.

For divisibility by 11, the sums of alternate digits need to either be equal, or differ by a multiple of 11, for example 1342 is divisible by 11 because the first and third digits add to the same number as the second and fourth digits.

Let’s consider when we have an odd number of digits. Then 1+k+9 would need to equal (or differ by 11n) 6+k. k in each case is some indeterminate number of 6s, but would cancel out in this case. 10 – 6 is 4, not a multiple of 11. But if the number of digits is even we have 1+k and k+9. The difference here is 8, again not a multiple of 11. So 166…9 can never be a multiple of 11.

Finally 169 is 13 x 13, and so 13 is the smallest number that divides a number from this sequence.