The total number of 6-digit numbers in which the sum of the digits is divisible by 5, is

We observe that first five places can be filled in 9 x 10 x 10 x 10 x 10 ways.

The sum of the digits in first five places can be of the following form:

5rn or Sm + 1 or Sm + 2 or Sm + 3 or Sm + 4 '

the possible values taken by the last digit corresponding to each of the possible values are shown below:

$\begin{array}{lc}\text{ Sum of the digits }& \text{ Last digit }\\ 5m& 0\text{ or }5\\ 5m+1& 4\text{ or }9\\ 5m+2& 3\text{ or }8\\ 5m+3& 2\text{ or }7\\ 5m+4& 1\text{ or }6\end{array}$

Thus, corresponding to each way of filling first five places there are two ways of filling the last digit.

Hence, required number of numbers= 9 x 10⁴ x 2 = 180000.