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?

The total number of ways are 144.

Step-by-step explanation:

Consider the provided information.

A number is divisible by 4 if the last two digits are divisible by 4.

Thus, the last two digit must be 04, 12, 20, 24, 32, 40, or 52.

If the last two digits are 04, 20, or 40, then rest of 3 digits can be: 4×3×2.

(It is given that no digit can repeat. If the last two digits are 04, 20, or 40, now we have only 4 digits for the first, 3 digits for the second and 2 digits for the third one.)

Therefore, the total number of cases are: (4×3×2)×3 = 72

If the last two digits are 12, 24, 32, or 52, then rest of 3 digits can be: 3×3×2.

Because 0 can't be the first digit.

Therefore, the total number of cases are: (3×3×2)×4 = 72

Hence, the total number of ways are 72+72=144

it should be a and d. lmk if i'm wrong xd

answer: ur v dumb

step-by-step explanation: