I have to use 4,5,6,7 and 8 only once and the divident can only be a 1 digit divident with no remainders and i can only use each digit once

Answer 1

To find a four-digit number that meets the given criteria, one possible solution is:

Start with the highest possible digit, which is 8. Use it as the thousands place.

Next, use the remaining four digits: 4, 5, 6, and 7.

To ensure there are no remainders, observe that the number needs to be divisible by 5. Hence, 5 should be in the units (ones) place.

Now, we have 8 _ _ 5 remaining.

To ensure there are no remainders, notice that the number needs to be divisible by 4. Hence, the leftmost two digits need to form a number divisible by 4. Among the remaining digits (6 and 7), the only combination that meets this requirement is 64. Thus, place 6 in the hundreds place and 4 in the tens place.

Putting it all together, the four-digit number that meets the given criteria is 8645.

Please note that there may be other possible solutions, and this is just one example.

Answer 2

The answer is 8645 divided by 7 equals 123

Answer 3

Apologies for the confusion in the previous response. Let's find a four-digit number using the digits 4, 5, 6, 7, and 8.

To ensure there are no remainders, we need to find a number that is divisible by 7. Let's try different combinations:
- 4, 5, 6, 7, 8: None of these numbers is divisible by 7.
- 5, 6, 7, 8: None of these numbers is divisible by 7.
- 4, 6, 7, 8: None of these numbers is divisible by 7.
- 4, 5, 7, 8: None of these numbers is divisible by 7.
- 4, 5, 6, 8: None of these numbers is divisible by 7.

It seems that there is no possible combination using these digits that results in a four-digit number divisible by 7 without a remainder.