I am confused with 1st option i.e m can't be odd in any case since m=96, when I factorise 96 then I get 2^5 * 3, therefore m would always be even.

m istelf doesn't equal 96; the product of its digits does. Since you've found an odd factor in its factorization, that factor could very well be the units digit!

Now, further explanation in case that doesn't unlock the problem:

Statement (1) tells us that m is odd, and we already know m is a 3-digit number. So it must end in 1,3,5,7, or 9. But it can't end in 5,7, or 9 because then there'd by no way to make the product of its digits 96 (since neither 5 nor 7 nor 9 goes into 96 cleanly). It also can't end in 1 because there'd be no way to get to the whole product of 96 from just the remaining two digits in that case (since the highest a digit can be is 9, and 9*9 only brings you to 81). So from Statement (1) we glean that the last (units) digit must be 3.

Ashley@VeritasPrep wrote:

I am confused with 1st option i.e m can't be odd in any case since m=96, when I factorise 96 then I get 2^5 * 3, therefore m would always be even.

m istelf doesn't equal 96; the product of its digits does. Since you've found an odd factor in its factorization, that factor could very well be the units digit!

yes I agree that if 3 is in unit's place then my unit place becomes odd but I have to get m as 96 and for that I need to have Tenth and Hundredth digit as even, then tell how will m become odd if I have even on Hundredth and Tenth place and odd on one's place

ohhhh sorry yup I got it, I got confused with something else, anyways thanks a lot