MCQ
$n^2 - 1$ is divisible by $8$, if $n$ is:
- AAn integer.
- BA natural number.
- ✓An odd integer.
- DAn even integer.
✓
Answer
Correct option: C.
An odd integer.
Let a $= n^2 - 1$
Here $n$ can be even or odd.
Case $I: n =$ Even i.e., $n = 2k$, where $k$ is an integer.
$\Rightarrow a=(2 k)^2-1$
$\Rightarrow a=4 k^2-1$
At $k=-1,4(-1)^2-1=4-1=3$, which is not divisible by $8 .$
At $k=0, a=4(0)^2-1=0-1=-1$, which is not divisible by $8$ , which is not.
Case II: $n =$ Odd i.e., $n =2 k +1$, where $k$ is an odd integer.
$\Rightarrow a=2 k+1$
$\Rightarrow a=(2 k+1)^2-1$
$\Rightarrow a=4 k^2+4 k+1-1$
$\Rightarrow a=4 k^2+4 k$
$\Rightarrow a=4 k(k+1)$
At $k=-1, a=4(-1)(-1+1)=0$ which is divisible by $8 $.
At $k=0, a=4(0)(0+1)=4$ which is divisible by $8.$
At $k=1, a=4(1)(1+1)=8$ which is divisible by $8.$
Hence, we can conclude from above two cases, if $n$ is odd, then $n^2-1$ is divisible by $8 .$
Here $n$ can be even or odd.
Case $I: n =$ Even i.e., $n = 2k$, where $k$ is an integer.
$\Rightarrow a=(2 k)^2-1$
$\Rightarrow a=4 k^2-1$
At $k=-1,4(-1)^2-1=4-1=3$, which is not divisible by $8 .$
At $k=0, a=4(0)^2-1=0-1=-1$, which is not divisible by $8$ , which is not.
Case II: $n =$ Odd i.e., $n =2 k +1$, where $k$ is an odd integer.
$\Rightarrow a=2 k+1$
$\Rightarrow a=(2 k+1)^2-1$
$\Rightarrow a=4 k^2+4 k+1-1$
$\Rightarrow a=4 k^2+4 k$
$\Rightarrow a=4 k(k+1)$
At $k=-1, a=4(-1)(-1+1)=0$ which is divisible by $8 $.
At $k=0, a=4(0)(0+1)=4$ which is divisible by $8.$
At $k=1, a=4(1)(1+1)=8$ which is divisible by $8.$
Hence, we can conclude from above two cases, if $n$ is odd, then $n^2-1$ is divisible by $8 .$
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