Hello Vincen.

Let's take a look at your question.

If we divide both sides by 2^8 we can get $$\frac{2^{98}}{2^8}\ =\ \frac{256}{2^8}\cdot L\ +\ \frac{N}{2^8}\ \Leftrightarrow\ 2^{90}\ =\ L\ +\ \frac{N}{2^8}.$$ Now, we have that L+N/(2^8) must be an interger. The only option that satisfy this is N=0.

This is why the correct option is A.

I hope this explanation may help you.

Feel free to ask me again if you have a doubt.

Regards.