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 , where n is a non-negative integer, what is the greatest value of ?

- ½
- 1
- 5
- 32
- 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 . Therefore **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 is true for how many unique integer values of n, where n is a prime number?

- 7
- 4
- 2
- 1
- 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:

This is clearly true. Thus we have one instance.

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

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:

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.

Hi Chris,

I always got stuck with these types of problems..

would you please suggest me more practice problems like this?

I don’t know of any books or websites that focus just on exponent for the GRE. But I can recommend a few good books where you can go through tons of GRE Quant practice questions and pick out the exponent problems for practice.

The Quant books from the Manhattan GRE 8-book series can be helpful. You’re most likely to find exponent questions if you purchase the Manhattan GRE volume on number properties. But Manhattan’s book of GRE Quant word problems and their GRE QC & Data Interpretation volume may also have a significant number of problems that involve exponents in some way. Manhattan’s 5 lb. Book of GRE Practice Questions also has a nice wide range of Quant problems that you can pick and choose exponent questions from.

And of course, ETS’s selection of official practice math problems is always golden. For targeted practice with exponents, pick problems form the GRE Official Guide or the official book of GRE Quant practice questions.

thanks a lot

Hi Chris,

For question 2, is there another way to solve it? I read through the comments but I don’t get how you got to (4/5)^12.

Could we possibly just sub in different values for n until we get to 13?

Thanks

First, you could definitely sub different values for n until you arrive at (4/5)^13. However, this would require a lot of time consuming trial and error, and time is never on your side in GRE Quants.

Instead, it’s better to use shortcuts and estimation wherever possible, to get to the answer more quickly. This is the approach Chris took to the second, medium-hard level problem in this post.

I just re-read the comments myself, and I can understand why the comments alone might not be enough for you to understand the shortcuts Chris took. There hasn’t yet been an explanation of how this problem is solved, from the very first step to the very last one. But I can fix that! 🙂

So here are the steps in detail (and let me know if you still have questions after this, Allie).

STEP 1: Get rid of the negative signs in the exponential notations. To keep the numeric value of each expression the same, if you get rid of the negative sign in the exponents, you have to reverse the position of the numerators and denominators, “flipping” each fraction. And so,

(5/4)^-n < 16^-1becomes(4/5)^n <(1/16)^1, or simply(4/5)^n < 1/16.STEP 2: Use “number sense” (the ability to make hidden but relatively simple connections between numbers) to recognize that (4/5)^3 is 64/125, which is slightly more than 1/2.

STEP 3: Recognize that (1/2)^4 = 1/16, the lesser number in the inequality. This is also a matter of number sense, but it’s an easy thing to spot with a basic sense of numbers. 1^(anything) is still 1, and 2^4 is 16. You should be able to make this connection just by looking at 1/2 and 1/16 and recognizing that (1/2)^4 becomes 1/16. From there, you can realize that if (1/2)^4 = 16, and (4/5)^3 is slightly greater than 1/2, than (4/5)^3 must also be slightly greater than 1/16.

STEP 4: Safely guess that if (4/5)^3 is a little greater than 1/2, and (1/2)^4 is equal to 1/16, then (4/5)^(3*4)— or (45)6^12— is slightly greater than 1/16. Why do you take the ^4 from 1/2, and multiply it by the ^3 after 4/5? Well, remember that 1/2 is a modified, smaller approximation of the value of (4/5)^3. To get the real exponential value of 4/5 that slightly exceeds 1/16, you need to turn (1/2) back into (4/5)^3, again, and then add its ^4 exponent back in. The process looks like this: (1/2)^4 ===> [(4/5)^3]^4. And [(4/5)^3]^4 can also be expressed as (4/5)^3*4, or (4/5)^12.

STEP 5: Now you have an exponential expression that is just a tiny bit bigger than 1/16. And the root number of the exponent, 4/5, is less than 1. Anytime you exponentially multiply a number smaller than 1, it goes down in value. So if (4/5)^12 is slightly greater than 1/16, then (4/5)^13 will be slightly smaller than 1/16. This makes 13 the lowest value you can plug in for n in the original inequality.

These five steps may look complicated, as I’ve explained them in as much detail as possible. But if you can get the hang of this kind of estimation and number sense, approaching this problem with the steps above will be faster than simply using trial and error to arrive at 13 as the correct value for n.

Again, let me know if you have any more questions or doubts about this problem and how to solve it. 🙂

hi chris ,

In question .2 ,Is it (5/4^-n) or (5/4)^-n

thanks

Hi Adarsh 🙂

In question 2, the left side of the equation is (5/4)^(-n). To rewrite this term with a positive exponent, we need to “flip” the fraction, which leaves us with the reciprocal of the original term:

(5/4)^(-n) = (4/5)^(n)

Hope this helps 🙂

hi there ,

yeah i got that but it was written (5/4^-n).

thanks

Hi Adarsh 🙂 Happy to have helped! I included the parenthesis to emphasize what the expression was saying. Negative exponents or exponents with more than one term can look confusing based on the formatting of the exponent. On the actual exam, all exponents will be in superscript (and not expressed using the “^” symbol), so it should always be clear what exponent the base is being raised to 🙂

I still don’t understand number 1. Can you please explain using other words?

Hi Chris,

Would you be okay if you can explain why the answer is 13 for question #2, please?

So far, I understand (4/5)^n < 1/16 but from that point, I lost why the answer is 13.

Please let me know and I thank you in advance!:)

Hi Ji,

Sure, I’d definitely be happy to explain! This is quite a toughie 🙂

So, what I did with 4/5 is I realized that by making n = 3, I get 64/125. That is very close to 64/128. So though it’s a tad larger, 64/125 for convenience sake can be simplified to 1/2. At this point the problem can be rewritten, using a new variable for an exponent, r.

1/2^r = 1/16

In this case r would equal 4. Since (4/5)^3 = 1/2, we can say (4/5)^3(4) or 4/5^12 = 1/16. The thing is it doesn’t quite equal 1/16 since (4/5)^3 is slightly larger than 1/2. Therefore, (4/5)^12 will be slightly larger than 1/16. By increasing 12 by 1, to get 13, we will definitively make the left side of the equation smaller than 1/16.

Hope that helps!

Aha! Thanks to your step-by-step explanation, now it does make sense to me. However, I am pretty concerned that I won’t be able to get this type of question right on the actual test. Persistently to practice makes me perfect, I guess. 🙂

Thank you Chris! 🙂

Hello,

I have a question regarding the following method of solving Q3:

I basically converted the equation into:

n < (-2)^n < 135.43

After which I assumed n to be a positive prime leaving only 1 possible prime, ie. 2.

Is this alright? Or is there a flaw in this reasoning?

Thank you.

Please update the link for Question no. 2 above in the blog (medium difficulty problem) with the correct link posted by respondent/Magoosh member Chris on April 25, 2014 at 9:25 pm. The link posted by Chris lele or Magoosh is actually incorrect. Thanks for posting this problem.

Hi Samy,

That link seems to have disappeared :(. No worries–I wrote an entirely new question for the second question in the three above. Hope you like it!

<a (5/4^-n) < 16^-1

Could you please clarify question 2 above ? I don't see how a has to do with anything ?

Hi Laz,

Not sure how that got in there–sorry for any confusion! I’ve removed the “<a".

Hi!

This blog is super helpful! However, I live in a country where YouTube is banned so unfortunately I cant take advantage of the video explanations. Can you please at least tell me the answer to the ‘medium difficulty’ question. Thanks, I really appreciate it!

No problem 🙂

The answer is (A).

Hope that helps!

Hi Chris,

For your second question, I don’t think the link goes to the correct question. Section 5 question 12 is a word problem about floor space. Could you please provide the correct question?

Thanks,

Hanna

Hanna,

The link to the problem is: http://web.archive.org/web/20110605194015/http://www.ets.org/s/gre/pdf/practice_book_GRE_pb_revised_general_test.pdf

The link posted in this blog leads to another version of the test that does not contain the correct problem. It leads to the Second Edition of the test (which is still a good tool to utilize while studying.) However, the link in the description of the YouTube video posted in the blog leads to the version of the test that has the correct problem. I hope this helps.

Take care,

Christopher

Hi Chris,

Is that still the correct link for #2? Because when I click it and go to section 5, #12, it’s some rectangular area problem. Thanks!

Hmm…it’s still working for me…try it again, but if it doesn’t work let me know and I’ll try to figure something out :).

Chris,

I think Arpita is saying that at the section 5, the problem #12 does not match with the video you are posting.

On the page #76 there are no exponent problems that match your video, the same goes from the page #74 to #81; therefore, the post have the wrong page, section and problem number.

Hi Axl,

Hmm…that is strange…well, I can’t seem to find the problem either now. I will replace it with another exponent problem.

Thanks for point that out :)!

*just need a confirmation

The important thing i learned over here is that prime numbers cannot be negative integers, because if you put n=-2,-3,-5 etc, the equation above is satisfied for an infinite number of values, that’s why i answered “none of the above”

just as a confirmation, can negative integers be prime numbers?

Hi Shree,

Prime numbers are only positive integers. Negative integers cannot be prime numbers.

Hope that helps!

Everything is perfect just wanted to verify this below statement !

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

The answer is (C) i.e. 2 isn’t it ?

Actually, this is quite a tricky question :).

‘2’ is the only integer value that satisfies the equation. Thus there is a total of one integer value (the number ‘2’) that satisfies the equation.

Hope that makes more sense :).

thanks.. I was also confounded!

Hi Chris,

I am just coming with lot of questions, i hope u may not get frustrated! :).

The link you gave for second question in the ETS web, when i am searching for that answer i found there is an extra column called (P+), what is that P+ ?

Thanks.!

(P+) stands for the percentage of people who answered the question correctly. For instance, a difficult question only 25% or less of people get right. This can help you get a sense of the type of questions you are missing (meaning you be answering the easy ones correctly, and don’t get despaired if you miss the tough ones).

Hope that helps, and don’t hesitate to ask any more questions :).

Got all right ,last question logic as you said is 2 is only prime no which is even and we want that value of n which can make (-2)^n =+ve…

Good collection of mazed questions…

Thanks! There will be some more tough math questions coming soon :).

Oops! Prime numbers start from 2. How dumb of me 😛

Hi Syam, no worries 🙂

Chris,

Wouldnt ‘-2’ qualify as ‘unique’ integer? Does ‘unique’ imply that the absolute value should be unique?

Hi,

You said that “Thus the only instance in which the inequality holds true is if we plug in ‘2’, the answer is (D).” The answer is C then?

Thanks

Hey Zainab,

This is a tricky question – it asks for the number of instances. There is only one instance – when you plugged in the number 2.

Hi Chris,

Thank you for the post.

I found the last question very confusing the way -2^-n was written. How are you able to tell if it is -(2^-n) or (-2)^-n? Also, when there is no parenthesis displayed, wouldn’t you assume it is the former case since exponent has more priority?

Thanks.

Hi Peng,

You are right – that is ambiguous. The negative numbers should be in parentheses, e.g. (-2)^-2

Thanks for catching that!