Q:

How many four-digit numbers are there formed from the digits 1, 2, 3, 4, 5 (with possible repetition) that are evenly divisible by 4?

I am stuck on this problem.
I started to solve the problem by finding the set of all possible numbers that can be formed without concern about divisibility by $4$, that is 
$5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 = 5^{4}$
since if it divisible by $4$, it must be even number I must find the set of odd numbers that end with $1,3,5$ and exclude it from the main set, 
that is 
$5^{3}+5^{3}+5^{3}$
but I got stuck at the set that ends with $2,4$ since it is not guaranteed that those numbers must be divisible by $4$. 
Can anyone help me? 

A:

A number is divisible by four if its last two digits are divisible by four. The set of possible answers is:

**12
**24
**32
**44
**52

Where * is any of $\{1,2,3,4,5\}$. There are $5\cdot 5 = 25$ permutation options for each **. Thus there are $25 \times 5 = 125$ numbers divisible by four from the digits $\{1,2,3,4,5\}$.

