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?

For divisibility by 11, the alternating sum of digits must be divisible by 11. The number 4x5y790 has digits (left to right): 4, x, 5, y, 7, 9, 0. Alternating sum = (4 + 5 + 7 + 0) − (x + y + 9) = 16 − x − y − 9 = 7 − (x + y). For divisibility by 11, 7 − (x + y) must be a multiple of 11. So x + y = 7 (giving 0) or x + y = 18 (giving −11). When (x + y) = 14, remainder on division by 11 = 3. Checking: if x+y = 14, then 7 − 14 = −7, not a multiple of 11. The valid values: x+y = 7 gives remainder 7; x+y = 18 gives remainder 7. Re-examining the alternating sum convention from right: digits from right are 0,9,7,y,5,x,4. Sum at odd positions − sum at even positions = (0+7+5+4) − (9+y+x) = 16 − 9 − x − y = 7 − x − y. So x + y ≡ 7 (mod 11), meaning x + y = 7 or 18. Remainder when x+y is divided by 11 = 7 when x+y=7, or 7 when x+y=18. Hence the answer is 7.