If a, b, c and d are integers, is (2^a·3^b)/(2^c·3^d) even ?

Thanks for pointing out the mistake. I had read the question as a, b, c and d to be positive integers which is not so.
Since, a, b, c and d are integers we need to know their signs as well. So, E shall be the answer.

If a, b, c and d are integers, is \(\frac{2a∗3b}{2c∗3d }even?\)

\((1) a > c\\
(2) b = c + d\)

\(2^a-c * 3^b-d\)

1) a > c => a-c > 0

This means we have 2 with a positive exponent. But we still do not know the whether the exponent of 3 is positive or negative. If its positive then we will have an even integer. But if its negative then we will have a fraction.

2) b = c + d

\(2^a-c * 3^b-d = 2^a-c * 3^c+d-d = 2^a-c * 3^c\)

In this case we do not know the sign of exponent of neither 2 nor 3. If a-c is positive and C is positive then we will have an even integer. But if a-c is positive and c is negative then we will have a fraction.

3) a > c & b = c + d

By combining both statements we are left with following expression: 2^a-c * 3^c

We still do not know the sign of c. If c is positive then we will have an even integer. But if c is negative then we will have a fraction.

Therefore, its E.