The six-digit number $5ABB7A$ is a multiple of 33 for non-zero digits $A$ and $B$. Which of the following could be possible value of $A + B$?

## Concept & Logic For a number to be a multiple of 33, it must be completely divisible by its **co-prime factors**, which are $11$ and $3$. * **Divisibility by 11:** The difference between the sum of the digits in odd positions and the sum of the digits in even positions must be `0` or a multiple of `11`. * **Divisibility by 3:** The sum of all digits must be a multiple of `3`. ## Step-by-Step Solution **Given:** The number is $5ABB7A$. $A$ and $B$ are non-zero single digits. **Calculation / Deduction:** * **Step 1: Test for divisibility by 11.** Let's group the digits by alternating positions (from right to left): * Sum of odd positions (1st, 3rd, 5th): $A + B + A = 2A + B$ * Sum of even positions (2nd, 4th, 6th): $7 + B + 5 = 12 + B$ * Calculate the difference: $(12 + B) - (2A + B) = 12 - 2A$ For the number to be divisible by 11, this difference ($12 - 2A$) must be $0, 11, -11$, etc. Since $A$ is a non-zero digit, the only logical way this equals a multiple of 11 is if it equals $0$: $12 - 2A = 0 \implies 2A = 12 \implies A = 6$. * **Step 2: Test for divisibility by 3.** Now substitute $A = 6$ back into the number: $56BB76$. Sum of all digits: $5 + 6 + B + B + 7 + 6 = 24 + 2B$. This sum must be a multiple of 3. Let's test possible non-zero values for $B$: * If $B = 3$: Sum $= 24 + 6 = 30$ (Divisible by 3) * If $B = 6$: Sum $= 24 + 12 = 36$ (Divisible by 3) * If $B = 9$: Sum $= 24 + 18 = 42$ (Divisible by 3) * **Step 3: Find the matching sum ($A + B$).** We have three valid combinations for $(A, B)$: * $6 + 3 = 9$ * $6 + 6 = 12$ * $6 + 9 = 15$ Looking at the given options (8, 9, 10, 14), only $9$ is a listed possible value. ## Exam Strategy & Shortcut When formatting the alternating sum for the rule of 11, strategically align the equations so that variables cancel out. Notice how $B$ appears exactly once in both the odd and even position sets. When you subtract them, $B$ immediately cancels out completely ($+B - B = 0$), leaving an equation with only $A$. This lets you solve for $A = 6$ in a matter of seconds without any guesswork. ## Common Pitfall A frequent mistake is finding $A=6$ and assuming $B$ can be any number, thereby randomly picking an option. Because the question specifies "multiple of 33", ignoring the rule of 3 will lead to incorrect pairs like $A=6, B=4$ (which would make $A+B=10$, a trap option). You must verify both conditions. ## Final Answer **Therefore, the correct answer is 9.**