If x and y are two digits and the number 4x5y790 is divisible by 11, then what is the remainder, if x + y is divided by 11?

Correct Answer: (D) 7

For the number $4x5y790$ to be divisible by 11, the alternating sum of its digits must be a multiple of 11. Sum of digits at odd places from the right: $0 + 7 + 5 + 4 = 16$. Sum of digits at even places from the right: $9 + y + x = 9 + (x + y)$. The difference is $16 - (9 + x + y) = 7 - (x + y)$. For divisibility by 11, $7 - (x + y)$$\equiv$ 0 \pmod{11} $\iff$ x + y $\equiv$ 7 \pmod{11}. Because $x$ and $y$ are digits ($0$$\le$ x+y $\le$ 18), $x + y$ is either 7 or 18, both of which leave a remainder of 7 when divided by 11.