AntrAx problem

The result is surely a multiplication of 5. So, the last digit of it is either 0 or 5. But it cannot be a multiplication of 2, which makes 5 the only possible last digit of the number.

The result would be $1\times3\times5\dots\times99$

since it contains 5 as a factor and no even number is included, the last digit will always be $\boxed5$

Any odd no multiplied by any multiple of 5 always gives a number ending in 5...................

If suppose we start multiplying all the odd numbers starting from 1,3,5.....

As we reach 5 all the further products will be a multiple of 5,since there is no even numbers so the last digit can not be 0 so it will be 5.