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ATQ,
A number is divisible by 12 if it’s divisible by both 3 and 4.
Check last two digits ‘y4’.
‘y4’ divisible by 4 → y = 0, 2, 4, 6, 8
Maximum y = 8
Now check sum of digits:
3 + x + 7 + 2 + 5 + 8 + 4 = 29 + x
29 + x must be divisible by 3 → x = 1, 4, 7
Maximum x = 7
So, max (x + y) = 7 + 8 = 15
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