Fun with numbers! (Part 1)

for divisibility by 4 last 2 digit are divisible by 4 but since repetition is not allowed 00,44,88 can't be last two digits so we have 22 choice for last 2 digits ------Case-1 last 2 digits contain zero 04,08,20,40,60,80-6 choices--- for each choice total number possible=8C5 {factorial of 5} total=40320 ------Case-2 we have 16 choices----- for each choice total no possible=8C5 {factorial of 5}-7C4*{factorial of 4} total=94080 TOTAL=134400

Not sure if this is what Aryan Goyat meant, but this is what I did:

The last two digits must be a multiple of 4. There are 24 positive multiples of 4 with less than 3 digits (the single-digit multiples 4 and 8 can be written as 04 and 08). There are two multiples with repeating digits, 44 and 88, but the rest all have distinct digits. Therefore, there are 22 possibilities for the last two digits of the number. Consider two cases:

Case 1: The last two digits have a zero.

Let the last two digits be a multiple of 4 that has the digit zero. There are $P(8,5)$ ways to select the first five digits of the number. Since there are six multiples of 4 with zero as a digit, there are $P(8,5)\times6$ numbers for this case.

Case 2: The last two digits do not have a zero.

Let the last two digits be a multiple of 4 that does not have the digit zero. There are 7 options for the first digit because it cannot be zero. Then, there are $P(7,4)$ ways to choose the second through fifth digits. There are $22-6=16$ multiples of 4 without zero as a digit, so there are $7\times P(7,4)\times16$ numbers for this case.

The two cases are disjoint (the last two digits either have a zero or they don't) and cover all possible cases. Therefore, the total is $P(8,5)\times6+7\times P(7,4)\times16=\boxed{134400}$ .