It is Given that 2^32 + 1 is Exactly Divisible by a Certain Number GMAT Problem Solving

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Question: It is given that \(2^{32} + 1\) is exactly divisible by a certain number. Which one of the following is also divisible by the same number ?

\(2^{96} + 1\)
\(2^{16} - 1\)
\(2^{16} + 1\)
\(7 * 2^{33}\)
\(2^{64} + 1\)

Correct Answer: A

Solution and Explanation:

Approach Solution 1:

The problem statement states that \(2^{32}+1\) is exactly divisible by a certain number. It is asked to find out the expression that is also divisible by the same number.
As per the formula of the mathematical rule:
a³ + b³ =(a+b)(a² -ab +b²)
Now, let’s assume that \(2^{32}+1\) is (a+b)
Then, a³ + b³ = \(2^{96}+1\)
As noted in the formula above,
a³ + b³ is always divisible by (a+b)
Therefore, any factor of (a+b) is a factor of (a³ + b³)
The expression that is also divisible by the same number = \(2^{96}+1\)

Approach Solution 2:

The problem statement states that \(2^{32}+1\) is exactly divisible by a certain number. It is asked to find out the expression that is also divisible by the same number.
Let’s solve the question by analysing the options.

a.\(2^{96}+1\)

We can rewrite \(2^{96}+1\) in the following way:

=> \({2^{96}}+1\) = \(({2^{32}})^3\) + \((1)^3\)
=>\({2^{96}}+1\) = \(({2^{32}+1})[({2}^{32})^2+1^2-{2}^{32}]\)
=> \({2^{96}}+1\) = \(({2^{32}+1})[{2}^{64}+1^2-{2}^{32}]\)

Therefore, the given expression is divisible by \({2^{32}}+1\).
Since \({2^{32}}+1\) is a factor of \({2^{96}}+1\) therefore, any factor of \({2^{32}}+1\) will divide \({2^{96}}+1\)
Therefore, the expression that is also divisible by the same number = \({2^{96}}+1\)
Hence, option A satisfies the requirement of the question. So we can eliminate the other options.

Approach Solution 3:

The problem statement states that \({2^{32}}+1\) is exactly divisible by a certain number. It is asked to find out the expression that is also divisible by the same number.
Let us assume \({2^{32}}\) is x.
Then, we can say \({2^{32}}+1\) = (x + 1)
Let’s consider (x+1) to be completely divisible by the natural number N.
Then, we get:
\({2^{96}}+1\) = \([({2^{32}})^3+1]\)
= \((x^3+1)\)
= \((x+1)(x^2−x+1)\), which is completely divisible by N, since (x+1) is divisible by N.

Therefore, the expression that is also divisible by the same number = \({2}^{96} + 1\).

“It is given that \({2}^{32} + 1\) is exactly divisible by a certain number”- is a topic of the GMAT Quantitative reasoning section of the GMAT exam. This topic has been taken from the book “Kaplan GMAT Math Workbook”. The candidates must figure out each information of GMAT Problem Solving questions in order to solve the quantitative problems. The GMAT Quantitative section test candidates’ ability in dealing with quantitative problems. The GMAT Quant practice papers help the candidates to analyse varieties of questions that will enable them to polish up their mathematical skills and knowledge.

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*The article might have information for the previous academic years, please refer the official website of the exam.

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It is Given that 2^32 + 1 is Exactly Divisible by a Certain Number GMAT Problem Solving

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