If digits are allowed to be used more than once, you can form

4*5*3 =

There are 4 non-zero digits for the leading digit, and 3 odd digits for the least-significant digit. The middle digit can be any of the five digits.

If

3*1*3 + 3*3*2 =

513, 813, 153, 853, 183, 583, 315, 815, 135, 835, 185, 385

4*5*3 =

**60 different numbers**.There are 4 non-zero digits for the leading digit, and 3 odd digits for the least-significant digit. The middle digit can be any of the five digits.

If

**each digit is only used once**, there are 3 digits for the least significant, 4 digits for the middle digit, and a different number of digits available for the most significant, depending on whether the middle digit is zero.3*1*3 + 3*3*2 =

**27 different numbers**301, 501, 801, 103, 503, 803, 105, 305, 805, 531, 831, 351, 851, 381, 581,513, 813, 153, 853, 183, 583, 315, 815, 135, 835, 185, 385