Question

If n is any odd number greater than 1, then $n^2 - 1$ is

• A. divisible by 96 always
• B. divisible by 48 always
• C. divisible by 24 always
• D. None of these

Hint:

For divisibility of $n^2 - 1$ with odd $n$, use factorization and check for multiples of 2 and 3 in consecutive evens.

Answer 1

The Correct Option is C

Step 1: Let $n$ be an odd number>1, e.g., $n = 3, 5, 7$. We need to check if $n^2 - 1$ is divisible by 24, 48, or 96.
Step 2: Factor $n^2 - 1 = (n - 1)(n + 1)$. Since $n$ is odd, $n - 1$ and $n + 1$ are consecutive even numbers.
Step 3: Consecutive even numbers include at least one multiple of 2.
Check for higher factors:
Among any three consecutive integers (e.g., $n-1, n, n+1$), one is divisible by 3.
Since $n$ is odd, $n-1$ and $n+1$ are even, and one of them is divisible by 4 (as every second even number is divisible by 4).
Step 4: Test divisibility by 8:
Among four consecutive even numbers, one is divisible by 8.
For $n$ odd, $n-1$ to $n+1$ span two evens, but extend to four: e.g., $n=3$, $2, 3, 4$, next even 6 includes 4 (divisible by 8).
Generally, $(n-1)(n+1)$ includes factors of 2 and 3.
Step 5: Check 24 = $8 \times 3$.
Test values: $n=3$, $3^2 - 1 = 8$,
divisible by 8, not 24. $n=5$, $25 - 1 = 24$,
divisible by 24. $n=7$, $49 - 1 = 48$,
divisible by 24. $n=9$, $81 - 1 = 80$, divisible by 24.
Step 6: Since $n$ is odd,
$n-1$ and $n+1$ are even, and their product is divisible by 4 (two evens) and 6 (one divisible by 3),
hence by 24. Thus, (C) is correct.