The 3-digit number 4a3 added to 984 gives 13b7, which is divisible by 11. Find (a + b).

Given data

• 4a3 + 984 = 13b7 (four-digit result)
• 13b7 is divisible by 11

Concept / Approach

• Use column-wise addition with carries.
• Apply divisibility rule for 11: (sum of digits in odd positions) − (sum in even positions) must be a multiple of 11 (i.e., 0, ±11, …).

Step-by-step calculation

Ones: 3 + 4 = 7 ⇒ matches the ones digit 7, carry = 0.Tens: a + 8 (+0) = b. For now, b = a + 8, with carry_tens = 0 (since a ∈ {0,1} to keep sum < 10; see next line).Hundreds: 4 + 9 + carry_tens must produce hundreds digit 3 with a carry 1 to thousands. So 4 + 9 + carry_tens = 13 ⇒ carry_tens = 0, hence a ∈ {0,1} and b = a + 8.Check a = 0 ⇒ 13b7 = 1387. Divisibility by 11: (1 + 8) − (3 + 7) = 9 − 10 = −1 ≠ 0 ⇒ reject.Check a = 1 ⇒ b = 9 ⇒ 13b7 = 1397. Divisibility by 11: (1 + 9) − (3 + 7) = 10 − 10 = 0 ⇒ divisible.

Result

a = 1, b = 9 ⇒ a + b = 10.

Verification

413 + 984 = 1397 ✓ and 1397 is a multiple of 11.

Common pitfalls

• Applying the divisibility-by-11 rule incorrectly (ensure correct positional sums).
• Forgetting that the tens column must not exceed 9 to keep the carry 0 as required by the hundreds column constraint.

Final Answer

10