Here is the pattern of the ones digit: 7 9 3 1 7 9 3 1 7 9 3 1

(1) x is a multiple of 6

Let's say x = 6 and y =1

then it would be 87^6 and the ones digit would be 9

Let's say x = 6 and y = 2

then it would be 87^12 and the ones digit would be 1

Insuff

(2) y is a multiple of 2
if y=2 and x=3

then it would be 87^6 and the ones digit would be 9

if y =4 and x=3

then it would be 87^12 and the ones digit would be 1

Insuff

Together

if x=6 and y=2

then it would be 87^12 and the ones digit would be 1

if x=6 and y=4

then it would be 87^24 and the ones digit would be 1

if x=12 and y=3

then it would be 87^36 and the ones digit would be 1

I will go with C

If x and y are positive integers and x+y>3, then what is the units digit of 87^(x^y)?

(1) x is a multiple of 6
(2) y is a multiple of 2


Statements A and B together are sufficient,however,the statements on their own are not.

The digit 7 follows a cyclic pattern of 4.
Now 6=283 and hence 2 is a factor of x.now if y has 2 as a factor it means y is even.Thus the power of e87 will always be evenly distributed by4 and the last digit will be 1.

Hence C