The 7-digit number 69A840B is divisible by 12. What is the maximum value of (A + B)?
15
A number is divisible by 12 if it is divisible by both 4 and 3. For divisibility by 4, the last two digits '0B' must form a multiple of 4, so B can be 0, 4 or 8.
For divisibility by 3, the digit sum 6 + 9 + A + 8 + 4 + 0 + B = 27 + A + B must be a multiple of 3, which means A + B must itself be a multiple of 3.
Testing B = 8: A + 8 is a multiple of 3, so A can be 1, 4 or 7; the largest gives A + B = 7 + 8 = 15. For B = 4 the best is A = 8, giving 12; for B = 0 the best is A = 9, giving 9. So the maximum of A + B is 15.
Hence, the correct answer is 15.
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Important Questions from Divisibility and Remainder
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