Question

Find the number of $4$-digit numbers (in base 10) having non-zero digits and which are divisible by $4$ but not by $8$.

Answer 1

We divide the even 4-digit numbers having non-zero digits into 4 classes;
those ending in $2,4,6,8$
$\left(A\right)$ Suppose a 4-digit number ends in $2$.
Then the second right digit must be odd in order to be divisible by $4$.
$\therefore$ the last $2$ digits must be of the form $12,32,52,72$ or $92$.
If a number ends in $12,52$ or $92$, then the previous digit must be even in order not to be divisible by $8$ and we have $4$ admissible even digits.
Now the left most digit of such a 4-digit number can be any non-zero digit and there are 9 such ways, and we get $9\times 4\times 3=108$ such numbers.
If a number ends in $32$ or $72$, then the previous digit must be odd in order not to be divisible by $8$ and we have $5$ admissible odd digit and there are $9$ such ways and we get $9\times 5\times 2=90$ such numbers.
$\therefore$ the number of 4-digit numbers having non-zero digits,ending in $2$, divisible by $4$ but not by $8$ is $108+90=198$
(B) If the number ends in $4$, then the previous digit must be even for divisibility by $4$.
Thus the last two digits must be of the form $24,44,54,84$.
If we take numbers ending with $24$ and $64$, then the previous digit must be odd for non-divisibility by $8$.
And the left most digit can take $9$ possible values. We thus get $9\times 4\times 2=72$ numbers.
$\therefore$ the admissible numbers ending in $4$ is $90+72=162$
$\left(C\right)$ If a number ends with $6$, then the last two digits must be of the form $16,36,56,76,96$.
For numbers ending with $16,56,76$, the previous digit must be odd.
For numbers ending with $36,76$, the previous digit must be even.
$\therefore (9\times 5\times 3)+(9\times 4\times 2)=135+72+207$ numbers.
$\left(D\right)$ If a number ends with $8$, then the last two digits must be of the form $28,48,68,88$.
For numbers ending with $28,68$, the previous digit must be even.
For numbers ending with $48,88$,the previous digit must be odd.
$\therefore (9\times 4\times 2)+(9\times 5\times 2)=72+90=162$ numbers.
$\therefore$ the number of 4-digit numbers , having non-zero digits and divisible by 4 but not by 8 is
$198+162+207+162=729$