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


Solution: We divide the even 4-digit numbers having non-zero digits into 4 classes: those ending in $2,4,6,8$.

(A) Suppose a 4-digit number ends in 2. Then the second right digit must be odd in order to be divisible by 4 . Thus 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 digits. Here again 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 $9 \times 5 \times 2=90$ such numbers. Thus 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 be any non-zero digit. Here we get $9 \times 5 \times 2=90$ such numbers. If the last two digits are of the form 44 and 84 , then previous digit must be even 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. Thus the admissible numbers ending in 4 is $90+72=162$.

(C) 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. Thus we get here $(9 \times 5 \times 3)+$ $(9 \times 4 \times 2)=135+72=207$ numbers.

(D) 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. Thus we get $(9 \times 4 \times 2)+(9 \times 5 \times 2)=$ $72+90=162$ numbers.

Thus the number of 4-digit numbers, having non-zero digits, and divisible by 4 but not by 8 is

$$
198+162+207+162=729
$$
