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

$\frac19$ of all six digit numbers start with a 5. So 100,000 six digit numbers are of the form 5******

This leaves 800,000 numbers that do not start with a 5.

$\frac1{10}$ of the remaining numbers have a 5 in the ten-thousands column, so we need to subtract 80,000 from 800,000 leaving 720,000.

$\frac1{10}$ of the remaining numbers have a 5 in the thousands column, so we need to subtract 72,000 from 720,000, leaving 648,000.

$\frac1{10}$ of the remaining numbers have a 5 in the hundreds column, so we need to subtract 64,800 from 648,000, leaving 583,200.

$\frac1{10}$ of the remaining numbers have a 5 in the tens column, so we need to subtract 58,320 from 583,200 leaving 524,880.

$\frac1{10}$ of the remaining numbers have a 5 in the units column, so we need to subtract 52,488 from 524,880, leaving 472,392.

A slightly quicker method would be to multiply by 0.9 instead of subtracting $\frac1{10}$ in each of the above steps.

Here is a different solution, 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 .