 ## Blog

### The Quest for 700: Weekly GMAT Challenge (Answer)

Yesterday, Manhattan GMAT posted a GMAT question on our blog. Today, they have followed up with the 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:

 Integer x 1 1/1 = 1 2 1/2 3 1/3 4 1/4 5 1/5

Now, just keep adding columns. Add one for y, defined as –x2:

 Integer x y = –x2 1 1/1 = 1 –1 2 1/2 –1/4 3 1/3 –1/9 4 1/4 –1/16 5 1/5 –1/25

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

Add one more column for xy. 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.

 Integer x y = –x2 xy 1 1/1 = 1 –1 1-1 = 1 2 1/2 –1/4 (1/2)^(–1/4) = 2^(1/4) 3 1/3 –1/9 (1/3)^(–1/9) = 3^(1/9) 4 1/4 –1/16 (1/4)^(–1/16) = 4^(1/16) 5 1/5 –1/25 (1/5)^(–1/25) = 5^(1/25)

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 22. The fourth answer becomes then (2^2)^(1/16) = 2^(1/8), which is less than 2^(1/4). The number that solves z8 = 2 is smaller than the one that solves z4 = 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.

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