how many multiples of 4 greater than 40000 but less than 70000 can be formed using the digits 0, 1, 3, 4, 6, 7, 8 if the repetition of digits is not allowed

The problem can be solved by considering the conditions for a number to be a multiple of 4 and the possible placements of the digits.

1. A number is a multiple of 4 if the number formed by its last two digits is a multiple of 4. This gives us the following possible pairs for the last two digits: 04, 08, 16, 32, 36, 60, 64, 68, 80, 84.

2. The number must be greater than 40000 but less than 70000, so the first digit can only be 4, 5, or 6.

3. The remaining three digits can be any of the remaining digits, without repetition.

Let's calculate the number of multiples for each possible first digit:

• If the first digit is 4, there are 10 possible pairs for the last two digits. For each of these pairs, there are 4 remaining digits for the second and third positions, which can be arranged in 43 = 12 ways. So, there are 1012 = 120 multiples of 4 starting with 4.

• If the first digit is 5, there are 10 possible pairs for the last two digits. However, we must exclude the pairs that include 4, because 4 is not available for the last two digits. This leaves us with 6 pairs. For each of these pairs, there are 4 remaining digits for the second and third positions, which can be arranged in 43 = 12 ways. So, there are 612 = 72 multiples of 4 starting with 5.

• If the first digit is 6, there are 10 possible pairs for the last two digits. However, we must exclude the pairs that include 4 and 5, because 4 and 5 are not available for the last two digits. This leaves us with 4 pairs. For each of these pairs, there are 4 remaining digits for the second and third positions, which can be arranged in 43 = 12 ways. So, there are 412 = 48 multiples of 4 starting with 6.

Adding up the multiples for each possible first digit, we get 120 + 72 + 48 = 240 multiples of 4 greater than 40000 but less than 70000 that can be formed using the digits 0, 1, 3, 4, 6, 7, 8 without repetition.