Question

For an odd positive integer $n$ ($51 \le n \le 99$), the quantity $n^3 - n$ is always divisible by:

• A. 48
• B. 24
• C. 18
• D. None of these

Hint:

For $n^3 - n$, factor to $n(n-1)(n+1)$ and analyze divisibility using properties of consecutive integers.

Answer 1

The Correct Option is A

To determine when \( n^3 - n \) is divisible, consider:

Expression: \( n^3 - n = (n)(n^2 - 1) = (n)(n-1)(n+1) \) 

Here, \( n \), \( n-1 \), and \( n+1 \) are three consecutive integers. This implies:

• Exactly one of these is divisible by 3.
• At least one is even, ensuring one is divisible by 2 and another by 4 (since one of every 3 consecutive numbers is divisible by 4 if they are odd like \( n \)).

Thus, the product \( n(n-1)(n+1) \) is divisible at least by \( 2 \times 4 \times 3 = 24 \).

Additionally: To check divisibility by 48, examine whether the product includes an extra factor of 2.

• Among the three values (\( n \), \( n-1 \), \( n+1 \)), two are even ensuring four 2's are present in the product. Hence divisibility by 48 (as \( 48 = 2^4 \times 3 \)) is confirmed:
  • Example check:
    \( n = 51 \rightarrow 51 \times 50 \times 52 = 132600 \equiv 0 \ (\text{mod } 48) \)

Conclusion: For all odd positive integers \( n \) where \( 51 \leq n \leq 99 \), \( n^3 - n \) is always divisible by 48.