Correct option is A
Given:
1. A = 162³ + 327³
2. B = 612³ - 123³
3. 489 = 3 × 163
Concept: For a number to be divisible by 489, it must be divisible by both 3 and 163. We will check divisibility for A and B separately.
Solution:
Step 1: Check Divisibility of A = 162³ + 327³
Divisibility by 3:
Both 162 and 327 are divisible by 3.
Their cubes, 162³ and 327³, are also divisible by 3.
Therefore, A = 162³ + 327³ is divisible by 3.
Divisibility by 163:
When 162 is divided by 163, the remainder is 162.
When 327 is divided by 163, the remainder is 1.
Substitute these into A: A = 162³ + 327³ becomes (-1)³ + 1³ modulo 163. (-1)³ is -1, and 1³ is 1. Adding these gives -1 + 1 = 0 modulo 163.
Since A is divisible by both 3 and 163, A is divisible by 489.
Step 2: Check Divisibility of B = 612³ - 123³
Divisibility by 3:
Both 612 and 123 are divisible by 3.
Their cubes, 612³ and 123³, are also divisible by 3.
Therefore, B = 612³ - 123³ is divisible by 3.
Divisibility by 163:
When 612 is divided by 163, the remainder is 123.
When 123 is divided by 163, the remainder is 123.
Substitute these into B: B = 612³ - 123³ becomes 123³ - 123³ modulo 163. This simplifies to 0 modulo 163.
Since B is divisible by both 3 and 163, B is divisible by 489.
Conclusion:
Both A and B are divisible by 489.
Answer: (a) Both A and B
