$2$ digit numbers that are divisible by 5

Question

I am trying to find all $2$ digit numbers that are divisible by $5$. So, if we put the $0$ at the last position - _$0$, we have $V_9^1=9$ options for the first digit. If we put $5$ at the last position - _$5$, we have $V_9^1=9$, but this includes $05$. The total number is: $2V_9^1-1=17.$ Aren't they $18$?

Answer 1

You have $9$ cases where the last digit ends with $5$ and $05$ is not one of them. Hence answer is $9+9=18$.

You can also divide the largest two-digit number $(95)$ by $5$ and subtract the number of one-digit multiples.

$={95\over 5}-{5\over 5} =19-1=18$

Answer 2

Two digits multiples of $5$ are $$10,15,20,....,95$$ That is $5$ times the elements of $${2,3,4,..,19}$$

Thus there are $18$ of them.