In this problem, we must do a permutation.
We'll answering this using something called the "counting principle" (probability).
Our 4-digit number can start with any digit that is not zero (if we assume examples like 0123 or 0009 are not 4 digit numbers, hence assume that leading 0s are not digits), further more, the other restriction in the number will be that it does not end in an even number or in 5; this means there are only 4 possible last digits in the number (i.e., 1,3,7, and 9).
Then, by the counting principle the answer is (9*10*10*4)=3600.
There are 4000 4-digit numbers that are not multiples of 2 nor 5.