Question: If \(2^{98}=256L+N\) , where L and N are integers and  \(0\leq{N}\leq{4}\), what is the value of N?

1. 1
2. 1
3. 2
4. 3
5. 4

“If \(2^{98}=256L+N\) , where L and N are integers and \(0\leq{N}\leq{4}\) , what is the value of N?”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Official Guide Quantitative Review”.
To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.

Answer

Approach Solution 1

Given: \(2^{98}=256L+N\)

Divide both parts by \(2^8:2^{90}=L+\frac{N}{2^8}\)

Now as both \(2^{90}\) and L are an integers then \(\frac{N}{2^8}\) must also be an integer, which is only possible for N = 0 (since \(0\leq{N}\leq{4}\) )

Correct option: A

Approach Solution 2

Given: \(2^{98}=256L+N\)

\(2^{98}=256L+N\). Now left hand expression is EVEN and we know E+E or O+O only gives output as EVEN

But since L is multiplied by \(2^8\) so ODD + ODD case is not possible.

Now, N can take only even values i.e., 0,2,4

If we put N = 2 then one 2 from N and L will cancel out from Right and Left side leaving below expression

\(2^{97}=2^7L+1\)

But then right side expression will give ODD value which is incorrect.

Same is the case with N=4.

So only possible value is 0

Correct option: A

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