Question

Using the digits from 0 to 9, how many 3-digit numbers can be constructed such that the number must be odd and greater than 500 and digits may be repeated?

Answer 1

`250` numbers

Explanation:

If the number is `ABC`, then:

For `A`, there are `9` possibilities: `5,6,7,8,9`

For `B`, all the digits are possible. There are `10`

For `C`, there are `5` possibilities. `1,3,5,7,9`

So the total number of `3`-digit numbers is:

`5xx10xx5 = 250`

This can also be explained as:

There are `1000,3`-digit numbers from `000 to 999`

Half of them are from `500 to 999` which means `500`.

Of those, half are odd and half are even.

Hence, `250` numbers.

Answer 2

250 numbers

Explanation:

The 1st digit must be greater than or equal to 5 for the number to be greater than 500. There are 5 possibilities (5, 6, 7, 8, 9).

The 2nd digit has no restriction on it. There are 10 possibilities (0-9).

The 3rd digit must be odd in order for the number to be odd. There are 5 possibilities (1, 3, 5, 7, 9).

`5*10*5=250` numbers