How many positive integers divisible by 4 can be formed using the digits 1, 2, 3 and 4, each at most once for each integer? How many positive integers divisible by 4 can be formed using the digits 1, 2, 3 and 4, each at most once for each integer?  Answer is 16. Can someone explain?
 A: A number is divisible by 4 if the number formed with its last two digits is divisible by 4. This implies numbers ending with 12, 24, 32. Therefore these numbers are 4, 12, 24, 32, 312, 412, 124, 324, 132, 432, 1324, 3124, 3412, 4312, 1432, 4132.
A: *

*Firstly, $4$ is such a number.

*For the number to be divisible by $4$, it must end with $12,24,32$.       $\therefore$  These $3$ numbers are also counted.

*For each of such case (number ending with $12,24,32$), if you want to form a three digit number, you have to put any one of the two remaining numbers in front of it . For example, if the ending is $12$, then you can put $3$ or $4$ in front of it forming $312$ and $412$. Thus, for $3$ cases we get $3 \times 2=6$ such numbers.

*If you form a four digit number, for each case you have two other numbers left and you can arrange them in $2!=2$ ways in front of it. For example,  if the ending is $12$, then you can put $34$ and $43$ in front of it forming $346$ and $436$. Thus, for three cases we have $3\times 2=6$ such numbers.


Summing up all the cases, we get $1+3+6+6=16$ ways which is the given answer.
