CBSE BoardEnglish MediumSTD 10MathsReal Numbers1 Mark
MCQ
$n^2- 1$ is divisible by $8,$ if $n$ is :
A
An integer.
B
A natural number.
✓
An odd integer.
D
An 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.
$\Rightarrow a = (2k)^2- 1$
$\Rightarrow a = 4k^2- 1$
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., $n = 2k + 1,$ where $k$ is an odd integer.
$\Rightarrow a = 2k + 1$
$\Rightarrow a = (2k + 1)^2- 1$
$\Rightarrow a = 4k^2+ 4k + 1 - 1$
$\Rightarrow a = 4k^2+ 4k$
$\Rightarrow a = 4k(k + 1)$
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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