The 7-digit number 12A916B is divisible by 24. What is the maximum value of (A + B)?
17
Divisible by 8 requires the last three digits '16B' to be divisible by 8: \(160+B\) must be divisible by 8, giving B=0 or B=8 (since 160 is already divisible by 8).
Divisible by 3 requires the digit sum \(1+2+A+9+1+6+B = 19+A+B\) to be divisible by 3.
Testing B=8 (larger value): \(27+A\) must be divisible by 3, so A must be a multiple of 3; the maximum digit is A=9.
Check: number 1291968 has last three digits 968 (divisible by 8) and digit sum 36 (divisible by 3) — both conditions hold.
Maximum \(A+B = 9+8 = 17\).
Hence, the maximum value of (A+B) is 17.
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