We have number n = a * (10 ^ n) + b.

n mod a = 0
and
n mod b = 0

If n mod a = 0 that b mod a = 0.
If n mod b = 0 that a * (10 ^ n) mod b = 0

We have now
b mod a = 0
a * (10 ^ n) mod b = 0

Ok, now let's imagine b = k * a. First condition is fulfilled.
a * (10 ^ n) / (k * a) = 10 ^ n / k

K is divider of 10 ^ n.
Let's note that k is less then 10. Why? Because a * 10 have n + 1 digit but b must have n digit.
Now for all 1 <= k <= 9 and 10^n mod k = 0 we calculate how many good numbers a we can use, a * k must have n digit that 10 ^ (n - 1) <= a <= (10 ^ n - 1) // k. Sum of all good ways to choose a and k is answer!