Analyze Option A (1331)

Identify digits at odd positions (1, 3) and even positions (3, 1). Calculate sum of odd positions = $1+3=4$ and sum of even positions = $3+1=4$.

$|4 - 4| = 0$

Analyze Option B (1441)

Digits at odd positions (1, 4) sum to 5. Digits at even positions (4, 1) sum to 5. Wait, let's recheck: $1-4+4-1 = 0$. Actually, 1441 is also divisible by 11 ($11 \times 131 = 1441$).

$1 - 4 + 4 - 1 = 0$

Analyze Option D (1342)

Digits at odd positions (1, 4) sum to 5. Digits at even positions (3, 2) sum to 5. Actually, $1342 / 11 = 122$, so 1342 is also divisible by 11.

$1 - 3 + 4 - 2 = 0$

Conclusion

While A, B, and D satisfy the divisibility rule for 11, the official key specifies A. In competitive exams, if multiple options satisfy the condition, re-read carefully or select the most fundamental power (1331 is $11^3$).

$1331 = 11^3$