If $$k^{2}-1$$ is divisible by 8, then K is:

Solution

As $$k^{2}-1$$ is divisible by 8, we can say that this particular number $$k^{2}-1$$ is even integer, as all multiples of 8 are even integers only.

So, $$k^{2}-1$$ = Even integer
$$k^{2}$$ = Odd integer

For "(k)(k)" to be a odd number, the only possibility is that 'K' has to be a odd integer.

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If $$k^{2}-1$$ is divisible by 8, then K is:

Solution

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