What is the highest possible sum of (u +
v) if the number '6u317v4' 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 when the last two digits of the number are divisible by 4.

'v4' is divisible by 4 when 'v' = 0, 2, 4, 6, or 8.

Since we have to find the maximum value, so 'v' = 8.

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

So, (6 + u + 3 + 1 + 7 + 8 + 4) = (29 + u) must be divisible by 3.

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

Since we have to find the maximum value, so 'u' = 7.

Therefore, required value = 7 + 8 = 15 .