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ATQ,
A number is divisible by 12 if it is divisible by both 3 and 4.
A number is divisible by 4 when the last two digits of the number are divisible by 4.
'y6' is divisible by 4 when 'y' = 0, 2, 4, 6, or 8.
Since we have to find the maximum value, so 'y' = 8.
A number is divisible by 3 when the sum of its digits is divisible by 3.
So, (7 + x + 5 + 4 + 2 + 8 + 6) = (32 + x) must be divisible by 3.
So, 'x' = 1, 4, or 7.
Since we have to find the maximum value, so 'x' = 7.
Therefore, required value = 7 + 8 = 15 .
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