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

Sagot :

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).


[tex] So for 10^4 -> 10^4 - 9^4 = 10.000 - 6.561 = 3.439 [/tex]

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

Calculistaidn

we know that

It is much easier to compute the number of numbers that do not contain a [tex] 3 [/tex].

In the first [tex] 10 [/tex] numbers [tex] 9 [/tex] do not contain a [tex] 3 [/tex].

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

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

[tex] 9*9 = 81 ways. [/tex]

In the first [tex] 1000 [/tex] numbers, reasoning as above, there are

[tex] 9^{3} [/tex] numbers that do not contain a [tex] 3 [/tex].

Now the pattern should be becoming obvious.

In the first [tex] 10^{n} [/tex] numbers there are [tex] 9^{n} [/tex] numbers that do not contain a [tex] 3 [/tex].

So

In the first [tex] 10^{4} [/tex] numbers there are [tex] 9^{4} [/tex] numbers that do not contain a [tex] 3 [/tex].

[tex] 9^{4}= 6,561 [/tex]

[tex] 10,000- 6,561= 3,439[/tex]

Thus

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

[tex] Ratio=\frac{3,439}{10,000} \\ \\ Ratio=0.3439 [/tex]

therefore

the answer is

[tex] 0.3439 [/tex]

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