Find the highest value of (x +
y) if '7x482y6' is divisible by 12.

ATQ,

A number is divisible by 12 if it is divisible by both 3 and 4.

A number is divisible by 4 if the last two digits of the number form a number divisible by 4.

'y6' is divisible by 4 when 'y' = 0, 2, 4, 6 or 8.

Since we need the highest value, so 'y' = 8.

A number is divisible by 3 if the sum of all its digits is divisible by 3.

(7 + x + 4 + 8 + 2 + 8 + 6) = (35 + x) must be divisible by 3.

So, 'x' = 1, 4 or 7.

Since we need the highest value, so 'x' = 7.

Thus, required value = 7 + 8 = 15.