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If 298=256L+N298=256L+N, where LL and NN are . . . . . .

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by EconomistGMATTutor » Wed Jan 03, 2018 1:23 pm
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.
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