n2 – 1 is divisible by 8, if n is ………………(A) an odd numbe

1.	n2 – 1 is divisible by 8, if n is ………………(A) an odd number (B) an even number (C) prime number (D) integer
Answer» Correct option is (A) an odd number If n = 2 then \(n^2-1=2^2-1\) = 4 - 1 = 3 which is not divisible by 8. Thus, options B, C and D are incorrect. Hence, only A can be correct. Alternative :- Let n = 2m+1, \(m\in N.\) Then \(n^2-1=(2m+1)^2-1\) \(=4m^2+4m+1-1\) \(=4m^2+4m\) = 4m (m+1) \(\because\) m (m+1) is always even number. Thus m (m+1) = 2p, \(p\in N\) \(\therefore\) \(n^2-1\) = 4m (m+1) = 8p, \(p\in N\) \(\therefore\) \(n^2-1\) is divisible by 8 if n is an odd number. Correct option is (A) an odd number

n2 – 1 is divisible by 8, if n is ………………(A) an odd number (B) an even number (C) prime number (D) integer

Answer»

Correct option is (A) an odd number

If n = 2 then \(n^2-1=2^2-1\) = 4 - 1 = 3 which is not divisible by 8.

Thus, options B, C and D are incorrect.

Hence, only A can be correct.

Alternative :-

Let n = 2m+1, \(m\in N.\)

Then \(n^2-1=(2m+1)^2-1\) \(=4m^2+4m+1-1\)

\(=4m^2+4m\) = 4m (m+1)

\(\because\) m (m+1) is always even number.

Thus m (m+1) = 2p, \(p\in N\)

\(\therefore\) \(n^2-1\) = 4m (m+1) = 8p, \(p\in N\)

\(\therefore\) \(n^2-1\) is divisible by 8 if n is an odd number.

Correct option is (A) an odd number