How many 5 digit numbers can be formed using the digits 0 2 3 4 5 when repetition is allowed that the number is divisiible by 2 or 5 or both?

Answer 1

In order to be divisible by both 2 and 5, the number must be divisible by 10.

That pretty well means that the last digit always has to be zero.

We're also going to assume that since each number must be a 5-digit, leading zeros

are ruled out.

First digit . . . . . 2, 3, 4, 5 . . . . . 4 possibilities

Second digit. . . 0, 2, 3, 4, 5 . . . 5 possibilities

Third digit. . . . . 0, 2, 3, 4, 5 . . . 5

Fourth digit . . . 0, 2, 3, 4, 5 . . . 5

Fifth digit . . . . . 0

Total number of possibilities = 4 x 5 x 5 x 5 = 500