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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.
'q3' is divisible by 4 when 'q' = 0, 2, 4, 6, or 8.
Since we have to find the maximum value, so 'q' = 8.
A number is divisible by 3 when the sum of its digits is divisible by 3.
So, (4 + p + 6 + 7 + 9 + 8 + 3) = (37 + p) must be divisible by 3.
So, 'p' = 2, 5, or 8.
Since we have to find the maximum value, so 'p' = 8.
Therefore, required value = 8 + 8 = 16 .
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