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Manhattan GMAT Challenge Problem of the Week – 4 June 2012

by Manhattan GMAT on June 4th, 2012
2 comments
Click here to read more articles from Manhattan GMAT and to learn more about Manhattan GMAT's classes.
Posted in
  • Arithmetic
  • GMAT Math
  • Powers and Roots
  • Problem Solving

Here is a new Challenge Problem! If you want to win prizes, try entering our Challenge Problem Showdown. The more people that enter our challenge, the better the prizes!

Question

If x is the reciprocal of a positive integer, then the maximum value of x^y, where y = –x^2, is achieved when x is the reciprocal of

A. 1
B. 2
C. 3
D. 4
E. 5

Answer

As in so many of these problems, the first task is careful reading. Note that x is not itself the integer (except in the case of 1/1); x is the reciprocal of an integer. You might even make a quick table corresponding to the values in the answer choices:

Now, just keep adding columns. Add one for y, defined as –x^2:

Notice that according to PEMDAS, you apply the squaring operation (Exponent) before the negative sign (Subtraction).

Add one more column for x^y. For clarity, we’ll use the caret symbol (^) to indicate exponents. Notice that the negative sign undoes the reciprocal in x, leaving you with an integer base.

Finally, you are left with the task: which of those numbers in the last column is largest? Remember, fractional exponents are roots, so the second number is the fourth root of 2, and so on.

You can quickly eliminate 1, since the positive roots (to any degree) of any integer greater than 1 are always greater than 1. For instance, what is the 25th root of 5? It must be a number bigger than 1, even if only slightly, because that number times itself 25 times in all produces 5. So you can eliminate A.

Let’s compare the second and fourth numbers, because the base of 4 can be rewritten as 2^2. The fourth answer becomes then (2^2)^(1/16)=2^(1/8), which is less than 2^(1/4). The number that solves z^8 = 2 is smaller than the one that solves z^4=2. So the answer can’t be D.

What about B versus C? Raise both numbers to the 36th power, so that we eliminate fractional exponents.

2^(1/4)^36=2^9

3^(1/9)^36=3^4

Which of the results is bigger? 2^9=512, while 3^4= only 81. So the original numbers must be in that same order of size; B is bigger than C. C is out.

A similar argument takes out E (raise both B and E to the 100th power):

2^(1/4)^100=2^25

5^(1/25)^100=5^4= 125 = much smaller than 2^25.

The correct answer is B.

Special Announcement: If you want to win prizes for answering our Challenge Problems, try entering our Challenge Problem Showdown. Each week, we draw a winner from all the correct answers. The winner receives a number of our our Strategy Guides. The more people enter, the better the prize. Provided the winner gives consent, we will post his or her name on our Facebook page.

If you liked this article, let Manhattan GMAT know by clicking Like.

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2 comments

  • harsh Singla on June 5th, 2012 at 12:17 am

    This was really a nice questions.. :)

    Reply to this comment
  • rs on June 10th, 2012 at 7:59 am

    can you explain in detail why 1 and 5 are eliminated? 

    Reply to this comment

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