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

For divisibility by 11, the alternating sum of digits of 4x5y790 must be a multiple of 11. Digits from left: 4, x, 5, y, 7, 9, 0. Alternating sum = 4 - x + 5 - y + 7 - 9 + 0 = 7 - x - y. For this to be divisible by 11: 7 - x - y = 0, 11, -11, etc. Since x, y are digits (0-9), x + y ranges 0 to 18. 7 - x - y = 0 → x + y = 7. 7 - x - y = -11 → x + y = 18. 7 - x - y = 11 → x + y = -4 (impossible). So x + y = 7 or x + y = 18. Remainder when 7 is divided by 11 = 7; remainder when 18 is divided by 11 = 7. Both give remainder 7. Hmm — but the answer choices include 3. Let me reconsider alternating sum convention: (sum of digits at odd positions) - (sum at even positions). Positions 1,3,5,7: 4+5+7+0=16. Positions 2,4,6: x+y+9. Difference = 16 - (x+y+9) = 7 - x - y. Same result. So x+y = 7 or 18, remainder = 7. Answer (d) 7.