The 10-digit number 79x00001y6 is exactly divisible by 88. What is the value of (x + y)?

79x00001y6 is exactly divisible by 88 = 8*11 then it's divisible by both 8 and 11.

Next, a number is divisible by 11 if the difference between the sum of the odd placed digits (1st, 3rd, 5th...) and the sum of the even placed digits (2nd, 4th...) must be 0 or divisible by 11. 

So, (9 + 0 + 0 + 1 + 6) - (7 + x + 0 + 0 + y ) = 0

i.e; x + y + 7 = 16

i.e; x + y = 9

A number is divisible by 8 only if its last three digits must be divisible by 8, so 1y6 must be divisible by 8. 

Possible value of y = 3 , 7

When y = 7 , then x = 2 

When y = 3 , then x = 6 

x + y = 

6 + 3 = 9 ( condition-1)

7 + 2 = 9 (condition-2)