The 7-digit number 46A484B is divisible by 24. What is the minimum value of (A + B)?
1
Since \(24 = 8 \times 3\) and 8 and 3 are coprime, the number must be divisible by both 8 and 3.
Divisibility by 8 depends on the last three digits, which are \(84B\). Testing values of B, \(840 \div 8 = 105\) works, so \(B = 0\) is the smallest valid choice.
Divisibility by 3 requires the digit sum to be a multiple of 3. The digit sum is \(4 + 6 + A + 4 + 8 + 4 + B = 26 + A + B\).
With \(B = 0\), we need \(26 + A\) to be a multiple of 3. The smallest such A is \(A = 1\), since \(26 + 1 = 27\) is divisible by 3.
This gives \(A + B = 1 + 0 = 1\), and the number 4614840 is indeed divisible by 24.
Hence, the minimum value of \(A + B\) is 1.
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