Divide in Moderation.

we know that if the last two digit of a number is divided by 4, then that number is divided by 4... $\\$ so the numbers should have last two digits which are odd multiple of 4(because it shouldn't be divisible by 8) $\\$ those two digit numbers are 04, 12, 20, 28, 36, 44, 52, 60, 68, 76, 84, 92.. take the non-zero numbers [12, 28, 36, 44, 52, 68, 76, 84, 92] (9 numbers possible) $\\$ now the number of possible first two digits are $9^{2}$ ... $\\$ $\\$ so the total numbers $9^{3}$ $\\$ = $\boxed{729}$

I have posted here two possible solutions ; and the second one is shorter , so you may directly view that if you find the first one too lengthy:

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 × 4 × 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 × 5 × 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× 5× 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 × 4 × 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 × 5 × 3) + (9 × 4 × 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 × 4 × 2) + (9 × 5 × 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 *.

Alternative and Shorter Solution :. If we take any four consecutive even numbers and divide them by 8, we get remainders 0,2,4,6 in some order. Thus there is only one number of the form 8k + 4 among them which is divisible by 4 but not by 8. Hence if we take four even consecutive numbers

1000a + 100b + 10c + 2,

1000a + 100b + 10c + 4,

1000a + 100b + 10c + 6,

1000a + 100b + 10c + 8,

there is exactly one among these four which is divisible by 4 but not by 8. Now we can divide the set of all 4-digit even numbers with non-zero digits into groups of 4 such two consecutive even numbers with a, b, c nonzero. And in each group, there is exactly one number which is divisible by 4 but not by 8. The number of such groups is precisely equal to 9 × 9 × 9 = 729 , since we can vary a, b.c in the set {1, 2, 3, 4, 5, 6, 7, 8, 9}.