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ATQ,
Divisibility by 8 → Last 3 digits: y52. Try y = 1: 152 ÷ 8 = 19 (remainder 0), so divisible.
Now check divisibility by 11:
Alternating sum = 7 - x + 3 - 1 + 2 - 6 + 5 - 1 + 2 = (7 + 3 + 2 + 5 + 2) - (x + 1 + 6 + 1) = 19 - (x + 8)
Let 19 - (x + 8) = 11 → x = 0
So x = 0, y = 1 → xy = 01 → √(01) = √1 = 1
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