Question: An integer between 1 and 300, inclusive, is chosen at random. What is the probability that the integer so chosen equals an integer to an exponent that is an integer greater than 1?

1. \(\frac{17}{300}\)
2. \(\frac{1}{15}\)
3. \(\frac{2}{25}\)
4. \(\frac{1}{10}\)
5. \(\frac{3}{25}\)

“An integer between 1 and 300, inclusive, is chosen at random. What is the probability that the integer so chosen equals an integer to an exponent that is an integer greater than 1?”- 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.

Solution and Explanation

Approach Solution (1):

Basically we need to find how many \(m^n\) (where n > 1) are between 1 and 300 inclusive
For n = 2, \(m^n\) < 300, m < 18, so there are 17 such numbers: \(1^2=1,2^2=4,3^2=9,4^2=16....,17^2=289;\)

For n = 3, \(m^3\)< 300, m < 7, so there are 6 such numbers \(1^3=1,2^3=8,3^3=27,4^3=64,5^3=125,6^3=216\). \(1^3=1=1^2\)and \(4^3=64=8^2\) have already been counted so, that leaves only 4 numbers;

Skip n = 4, since all perfect fourth power numbers are also perfect squares;
For n = 5, \(m^5\)< 300, m < 4, so there are 3 such numbers. Out these 3 numbers \(1^5=1=1^2\) , has already been counted so, that leaves only 2 numbers;
Skip n = 6 for the same reason as n = 3;
For n = 7, \(m^7\) < 300, m < 3, so there are 3 such numbers. Out these 3 numbers \(1^7=1=1^2\) has already been counted so, that leaves only 1 number.
Total = 17 + 4 + 2 + 1 = 24

The probability thus equals to \(\frac{24}{300}=\frac{2}{25}\)

Correct Answer: C

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