The 7-digit number 46A484B is divisible by 24. What is the minimum value of (A + B)?

Question

The 7-digit number 46A484B is divisible by 24. What is the minimum value of (A + B)?

1. A. 6
2. B. 3
3. C. 1 (Correct answer)
4. D. 4

Correct Answer

Option C — 1

Detailed Solution & Explanation

The correct answer is 1.

Shortcut Trick

• $24 = 8 \times 3$, so the number must satisfy both rules.
• Divisible by 8 — only the last three digits matter: $84B$. Testing, $840$ and $848$ both work, so $B = 0$ or $B = 8$.
• Divisible by 3 — the digit sum $4+6+A+4+8+4+B = 26 + A + B$ must be a multiple of 3.
• Take the smaller $B$ first: with $B = 0$, $26 + A$ must be divisible by 3, so $A = 1$ gives 27 ✓.
• Minimum $(A + B) = 1 + 0 = \textbf{1}$.

Alternate Method

• Check the other branch to confirm it cannot beat it: with $B = 8$, $34 + A$ must be a multiple of 3, so the smallest $A$ is 2, giving $A + B = 10$.
• Since $1 < 10$, the minimum is 1, with the number $4614840$.

Additional Information

• Divisibility rules worth carrying:
  • 8 — last three digits divisible by 8; 4 — last two; 2 — last digit even
  • 3 and 9 — digit sum divisible by 3 or 9
  • 11 — alternating digit sum divisible by 11
• For a composite divisor, split it into coprime factors and test each. Note $24 = 8 \times 3$ works because 8 and 3 are coprime; splitting as $4 \times 6$ would not be valid, since 4 and 6 share a factor.
• The question asks for the minimum of $A+B$, so both branches must be checked — stopping at the first valid pair is the common error.