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 $11$?
Detailed Explanation:
For the number 4x5y790 to be divisible by 11, apply the alternating digit sum rule: (sum of odd-position digits from right) − (sum of even-position digits from right) must be divisible by 11.
Odd positions (from right): 0, 7, 5, 4 → sum = 16. Even positions: 9, y, x → sum = x + y + 9. Thus, 16 − (x + y + 9) = 7 − (x + y) must be ≡ 0 (mod 11).
Since x, y are digits (0–9), x + y ∈ [0, 18], so 7 − (x + y) ∈ [−11, 7]. The only multiples of 11 in this range are 0 and −11. If 7 − (x + y) = 0 → x + y = 7; if 7 − (x + y) = −11 → x + y = 18. In both cases, (x + y) mod 11 = 7.
Therefore, the remainder when x + y is divided by 11 is 7, making Option 4 the correct answer.
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