Stuck with divisibility test in Permutations

Question

How many 5 digit numbers can be formed using digits 0 to 7, divisible by 4, if no digit occurs more than once in a number.

1. 1480
2. 780
3. 1360
4. 1240
5. None Of These

I could calculate the 5 digit numbers can be formed using digits 0 to 7 but I couldn't make out the divisible by 4 thing.

Answer 1

Hint: $abcde$ is divisible by $4$ if and only if either $d$ is odd and $e$ is divisible by $2$ but not $4$, or $d$ is even and $e$ is divisible by $4$.

Answer 2

A number is divisible by $4$ if and only if the number's last two digits are divisible by $4$. So, in this case (using digits 0 to 7, and no digit occurs more than once in a number), the number's last two digits have to be either of the followings : $$04,12,16,20,24,32,36,40,52,56,60,64.$$

So, if $0$ is included in the last two digits, then we can use $3$ numbers from $6$ numbers to make such number. So, we have $\frac{6!}{3!}=6\cdot 5\cdot 4$ cases.

If $0$ is not included in the last two digits, then the left most digit cannot be $0$, so we have $5\cdot 5\cdot 4$ cases.