How many different 4-digit numbers can be formed using the digits 2, 4, 5, 6, and 7 such that no digits repeat and the number is divisible by four?

A number is divisible by four if the last two digits are divisible by four.
The last two digits of a number divisible by four must be:
00, 04, 08, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96
You can only use the digits 2, 4, 5, 6, and 7, so eliminate from the numbers above all that contain the digits 0, 1, 3, 8, 9. Also digits can't repeat, so eliminate 44 too.
We are left with:
24, 52, 56, 64, 72, 76
There are 6 possibilities for the two last digits of the number.

You can use five digits and no digits can repeat, so when you use two of them for the last two digits, you can only use three digits.
For the first digit, you can use any of these three digits.
For the second digit, you can use any of these three digits except for the one you used for the first digit, so for the second digit you can use two digits.

To sum up, there are:
- 3 possibilities for the first digit
- 2 possibilities for the second digit
- 6 possibilities for the last two digits

Using the rule of product, you can calculate that there are 3×2×6=36 possibilities for the number.
You can form 36 four-digit numbers that satisfy these conditions.

If you divide 0.1 and 0.6 you wii get 0.166666 which rounds to .167 which is answer b