Solution:

We know that the last digit of an odd number is (1, 3, 5, 7, 9).

Assume there are three boxes.

Any of the nine digits can be entered in the first box (zero not allowed at first position)

As a result, the alternatives are numerous.  ⁹C₁

Any of the ten digits can be used in the second box.

As a result, the options are ¹⁰C₁.

Any of the five digits can be used in the third box (1,3,5,7,9)

As a result, the options are ⁵C₁.

As a result, the total number of outcomes is ⁹C₁ × ¹⁰C₁ × ⁵C₁ = 9 × 10 × 5 = 450