Got wrong...solution plz..

Here \(\frac{j^2}{k}\) is odd that is a given so the statement 'E/E can be odd or can be even ' is irrelevant.

\(\frac{j^2}{k}\) is an integer and that too odd. So what are values (even or odd) of k if j is even (given)

You have to go backwards it is already given that the fraction is an odd integer.
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Sandy
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Here let us consider the option one at a time:

Option A ::: j and k are both even: It may not be true because if j=1 and k=1 then we can get \(\frac{j^2}{k}= Odd\)

OPtion B::: j = k: it may not be true because if j=5 and k=1; it still satisfy the equation \(\frac{j^2}{k} = Odd\)

Option C ::: If j is even, k is even , this may be true \(\frac{j^2}{k} = odd\) or \(j^2\) = k*Odd, so if j = even then k has to be even to satisfy the equation (Even * Odd = Even)

Option D::: If j is divisible by k this may not be true because if j = 3 , k = 9 , it still satisfy the equation \(\frac{j^2}{k} = Odd\)

Option E ::: \(j^2 > k\) It may not be true, because if j = 1 and k = 1 then \(j^2 = k\) and it still satisfy the equation \(\frac{j^2}{k} = Odd\)
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Sippose j=4 and k=4 then ?it will be 16/4=4 how come C is the right answer