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GRE Exponents: Practice Question Set

Each of the math questions below is directly inspired by a question in the on-line Revised GRE test. I’ve provided an easier version of the question (#1) and a more difficult version of the question (#2).

My recommendation is to try the easier version first. Then, if you answer it correctly, click on the link, and take a stab at the actual Revised GRE question.

If you are able to answer that question correctly, then as prize – you get a fiendishly difficult question (#3). Okay, maybe that’s not a prize – but it is great practice for those aiming for the 90% on quant.

The good news is I have explanations. For the Revised GRE question, I have recorded an explanation video you can watch. Finally, it is a good idea to try the easy question before the medium one, and the medium question before the difficult one.

Good luck!

 

1. Difficulty: Easy

If 0<10^n < 1,000,000, where n is a non-negative integer, what is the greatest value of 1/2 ^n?

  1. ½
  2. 1
  3. 5
  4. 32
  5. 64

 

Explanation: Don’t think big – think small. That is the smaller n becomes the greater ½^n becomes. So what is the smallest value? You may be tempted to say 1, which would give us ½. But remember n = 0, because 10^0 = 1. Therefore 1/2^0 = 1 Answer: B.

The “hidden zero,” as I like to call it, is a classic GRE math trick. So always keep your eyes open, especially when you see “non-negative integer,” which includes zero.

 

2. Difficulty: Medium-Difficult

(5/4)-n < 16-1

What is the least integer value of n?

Explanation:

The best place to start here is by getting rid of the unseemly negative signs and translating the equation as follows:

(4/5)n < 1/16

A good little trick to learn using 4/5 taken to some power is that (4/5)3 = 64/125, which is slightly — but only slightly — greater than ½. Therefore, we can translate (4/5)3 to ½.

(1/2)4 = 1/16

That would make (4/5)12 a tad larger than 1/16. To make it less than 1/16 we would multiply by the final 4/5, giving us n = 13.

 

 

3. Difficulty: Hard

The equation n < 1/{(-2)^{-n}} < 135.43 is true for how many unique integer values of n, where n is a prime number?

  1. 7
  2. 4
  3. 2
  4. 1
  5. None of the above

This problem can be difficult, indeed downright inscrutable, unless you take your time and process one piece of information at a time. Once you understand what the problem is saying, you should be able to solve the question relatively quickly.

 

Explanation: 

The most important piece of info is n is a prime number. So do not start by plugging in zero or one. Neither is a prime. The lowest prime is 2. When we plug in ‘2’ we get:

2 < 1/{(-2)^{-2}} < 135.43

2 < 4 < 135.43

This is clearly true. Thus we have one instance.

 

As soon as we plug in other prime numbers a pattern emerges.

1/{(-2)^{-n}} is always a negative number if n is odd. Because all of the primes greater than 2 are odd, the number in the middle will always be negative:

1/{(-2)^{-3}}=-8

1/{(-2)^{-5}}=-32

Because in each case n is a positive number we can never have the middle of the dual inequality be positive, if n is an odd prime.

Thus the only instance in which the inequality holds true is if we plug in ‘2’, the answer is (D).

If you got that right – congratulate yourself. It’s a toughie.

 

Special Note:

For even more GRE questions, check out our GRE Quant problems with answers and explanations! And to find out where exponents sit in the “big picture” of GRE Quant, and what other Quant concepts you should study, check out our post entitled: What Kind of Math is on the GRE? Breakdown of Quant Concepts by Frequency.

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