No correct solution to this problem was origianlly received. However, Mary of Birchwood Community High School gave a sound argument that just needs some adaptation. Thank you Mary..

Firstly Mary considered the number of six digit numbers - this is 900,000.

10% of all six digit numbers start with a 5. So 90,000 six digit numbers are of the form 5******

This leaves 810,000 numbers that do not start with a 5. How many of these have a 5 as the second digit??

And so on.....

Here is a solution to this toughnut from Junwei of BHASVIC

Let the six digits number is abcdef, which a, b, c, d ,e, f represent a digit respectively.

For a, neither 0 nor 5 could place in it, thus, 8 digits are available here (1,2,3,4,6,7,8,9)

For b, c, d, e and f, they can't contain 5, hence, 9 digits are available for them (0,1,2,3,4,6,7,8,9)

Therefore, the no. of six digits number which does not contain any 5 is

8 * 9 * 9 * 9 * 9 *9 =472392 .