If 2^98=256L+N, where L and N are integers and 0 ≤ N ≤ 4, what is the value of N?
A) 0
B )1
C ) 2
D) 3
E) 4
The OA is the option A.
What is the best way to solve this PS question? Experts, can you give me your explanation here? Thanks in advanced.
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If 298=256L+N298=256L+N, where LL and NN are . . . . . .
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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.
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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