Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.
Sorry for my lateness and hi, chetan2u!
Thank you for your kind explanation!
Well....I would like to add some more because of my apology.
"q/r is an integer if and only if q is a multiple of r" (that is, it can be paraphrased as"q/r is not an integer if and only if q is not a multiple of r).
q/r is an integer --> q is a multiple of r
pf) Since q/r is an integer we have q/r = n, for some integer n.
Then we have q= nr --> q is a multiple of r.
q is a multiple of r ---> q/r is an integer
pf) Similarly since q is a multiple of r we have some integer n which satisfies q= nr. ---> q/r = (nr)/r = n which is an integer...
So, since 27 is a multiple of 3, 27/3=9 is an odd integer.
But as chetan2u pointed out it cannot be an odd integer(not even an integer) if we choose q and r as not multiple-factor relation.
---> for example, q= 27, r=5 ---> q/r = 27/5 is not even an integer.
We cannot say, therefore, that q/r MUST be an odd integer.
Cheers!