Question

What proportion of the first 10,000 natural numbers contain a 3

Answer 1

In the first 10 -> 10 - 9 = 1 (contain a 3).
In the first 100 -> 100 - 9 * 9 = 19 (contain a 3).
In the first 10^n -> 10^n - 9^n (contain a 3).



`So for 10^4 -> 10^4 - 9^4 = 10.000 - 6.561 = 3.439`

The answer is 3.439 numbers contain a 3 in the first 10.000

Answer 2

we know that

It is much easier to compute the number of numbers that do not contain a
`3`.

In the first
`10` numbers
`9` do not contain a
`3`.

In the first
`100` numbers
`81` do not contain a
`3`. (This is because these numbers are formed by selecting a digit out of

`[0, 1, 2, 4, 5, 6, 7, 8, 9]` for the first position and another digit out of the same set for the last position and this may be done in

`9*9 = 81 ways.`

In the first
`1000` numbers, reasoning as above, there are

`9^(3)` numbers that do not contain a
`3`.

Now the pattern should be becoming obvious.

In the first
`10^(n)` numbers there are
`9^(n)` numbers that do not contain a
`3`.

So

In the first
`10^(4)` numbers there are
`9^(4)` numbers that do not contain a
`3`.

`9^(4)= 6,561`

`10,000- 6,561= 3,439`

Thus

the ratio of those numbers that do contain a 3 over the total is equal to

`Ratio=(3,439)/(10,000) \\ \\ Ratio=0.3439`

therefore

the answer is

`0.3439`