stmt1,

When a is divided by c, the quotient is an odd integer.

No info about b.

If b is even then ab/c is even

If b is odd then ab/c is odd

Insuff

stmt2,

When b is divided by c, the quotient is an odd integer.

No info about a.

If a is even then ab/c is even

If a is odd then ab/c is odd

Insuff

Combining 1 and 2,

There are 2 scenarios in this case.

1. a=10,b=14,c=2

ab/c is even

2. a=21,b=9,c=3

ab/c is odd

Insuff

Pick E

should be E.
It is easy to see why I and II, independently are not sufficient.

For I & II:
6*10/2 = even
9*15/3 = odd

i wonder is it possible to solve some formal way
(1) a=c(2k+1) where k is integer
(2) b=c(2m+1) where m is integer
a*b/c=c*(2k+1)c(2m+1)/c=c*(2k+1)(2m+1) and now it depend on the value of c, if c is odd then all phrase is odd, but in the problem the value of c is not fixed so E is the answr
guys do you think this is right approach?

Great answers, everyone!

Clock60, you can certainly solve this one formally (and algebraically, in a way). Here's how I'd do it:

Statement 1 says that:

a/c = an odd integer

So...if you multiply both sides by c, you get:

a = c * odd

That means that, if c is even, so is a. But if c is odd, then a is odd. This statement is not sufficient.


Statement 2 says that:

b/c = an odd integer

b = c * odd

Similarly, this tells us that b and c are either both even or both odd, but we don't know which.


Taken together, we know that a = c * odd and that b = c * odd, so:

ab/c = [c(odd) * c(odd)] / c

We can divide out one of the cs in the numerator with the c in the denominator, but we still have c * odd * odd, meaning that if c is even, the whole number will be even, but that if c is odd the whole number will be odd. Therefore, we can algebraically demonstrate that neither statement is sufficient, even if taken together. The answer is E.

opps i thought an even/even always produces an odd result?? 10/2 = 5 and 6/2 = 3

Hey, HP:

But try 8/2 = 4. There aren't hard-and-fast even/odd rules for division, so you'll need to try those out for yourself.

Basically, even just means "divisible by 2". But to divide a number and still have a factor of 2, you have to have more evens in the numerator than in the denominator:

8/2 = 2*2*2/2 You'll cancel one two out of the numerator and still have some remaining

6/2 = 2*3/2 You'll cancel the only two out of the numerator and be left with only an odd remaining

Got it will be more diligent in my picking of values... Good problem thanks!