If \(2^{98}=256L+N\) , Where L and N are Integers and \(0 \leq N \leq 4\)  , What is the Value of N?

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. 0
2. 1
3. 2
4. 3
5. 4

This topic is a part of GMAT Quantitative reasoning section of GMAT. This question has been taken from the book "GMAT Quantitative Review" published in the year 2021. GMAT Quant section consists of a total of 31 questions. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT quant topics in the problem-solving part require calculative mathematical problems that should be solved with proper mathematical knowledge.

Answer:

Write the original equation: \(2^{98}=256L+N\) , where \(0 \leq N \leq 4\)

As we know that if we will multiply ‘2’ eight times then we will get the value 256.

This means \(2^8=256\)

Put this value in the original equation as written above, we will get:

\(2^{98}=2^8*L+N\)

Now, we will divide both the parts by \(2^8\) , the new equation will become:

\(\frac {2^{98}}{2^8}=\frac{2^8}{2^8}*L+\frac{N}{2^8}\)

We know that when dividing the terms with same bases and different exponents, you will subtract all the exponents.

Applying the same method in the above formed new equation, we will get:

\(2^{90}=L+\frac{N}{2^8}\)

Now as we know that \(2^{90}\) and L is integers (as given in question that L and N are integers), then this means \(\frac{N}{2^8}\) must also be an integer.

This is only possible when N = 0, since \(0 \leq N \leq 4\) .

Correct Answer: A

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