A 3-digit number $4a3$ is added to another 3-digit number $984$ to give the four-digit number $13b7$, which is divisible by $11$. Then, $(a + b)$ is

### Concept & Logic The divisibility rule of $11$ states that the difference between the sum of digits at odd places and the sum of digits at even places must be either $0$ or a multiple of $11$. ### Step-by-Step Solution * **Given:** * First number: $4a3$ * Second number: $984$ * Sum: $13b7$, which is divisible by $11$. * **Calculation / Deduction:** 1. Add the numbers to find the relation between $a$ and $b$: $$4a3 + 984 = 13b7$$ Looking at the columns: Units: $3 + 4 = 7$ Tens: $a + 8 = b \implies b - a = 8$ Hundreds: $4 + 9 = 13$ 2. Apply the divisibility rule of $11$ to $13b7$: Sum of digits at odd places (1st and 3rd from right): $7 + 3 = 10$ Sum of digits at even places (2nd and 4th from right): $b + 1$ Difference: $10 - (b + 1) = 9 - b$ 3. For $13b7$ to be divisible by $11$, the result $(9 - b)$ must be $0$ or a multiple of $11$. Since $b$ is a single digit ($0 \le b \le 9$), the only possible value is when $9 - b = 0$, giving $b = 9$. 4. Substitute $b = 9$ into our tens column equation: $a + 8 = 9 \implies a = 1$ 5. Calculate the required value $(a + b)$: $a + b = 1 + 9 = 10$ ### Exam Strategy & Shortcut You can quickly solve this by directly testing the divisibility rule on $13b7$. Sum of alternating digits must be equal: $1 + b = 3 + 7 \implies b = 9$. Once $b=9$, you know from the addition that $a+8=9 \implies a=1$. Their sum is $10$. This skips writing out the full addition columns and solves the problem in 10 seconds. ### Common Pitfall A common mistake is incorrectly evaluating the alternating sum from left-to-right versus right-to-left, or forgetting that $a$ and $b$ must be single digits between $0$ and $9$, which strictly limits the possible multiples of $11$ to just $0$. ### Final Answer Therefore, the correct answer is 10.