VJesus12 wrote:How many five-digit numbers can be formed from the digits 0, 1, 2, 3, 4, and 5, if no digits can repeat and the number must be divisible by 4?
A. 36
B. 48
C. 72
D. 96
E. 144
The OA is E.
I started making the list but this is too long. Experts, may you tell me an easy and faster way to get an answer?
To be divisible by 4, the last two digits of the number must be divisible by 4. Therefore, they can be 04, 12, 20, 24, 32, 40, and 52. We can split these into two groups: 1) 04, 20, 40, and 2) 12, 24, 32, 52
Group 1:
If the last two digits are 04, then there are 4 choices for the first (or ten-thousands) digit, 3 choices for the second (or thousands) digit, and 2 choices for the third (or hundreds) digit. So we have 4 x 3 x 2 = 24 such numbers if the last two digits are 04. Similarly, there should be 24 numbers if the last two digits are 20 or 40. So we have 24 x 3 = 72 numbers in this group.
Group 2:
If the last two digits are 12, then there are 3 choices for the first (or ten-thousands) digit (since it can't be 0), 3 choices for the second (or thousands) digit, and 2 choices for the third (or hundreds) digit. So we have 3 x 3 x 2 = 18 such numbers if the last two digits are 12. Similarly, there should be 18 numbers if the last two digits are 24, 32 or 52. So we have 18 x 4 = 72 numbers in this group, too.
Therefore, there are a total of 72 + 72 = 144 numbers.
Answer: E
