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