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GRE Math – Exponent Challenge

A relatively common exponent question on the GRE will ask you to figure out the units digit of some number raised to an exponent, which usually results in a number roughly equivalent to the number of grains of sand in the Sahara, and thus out of reach of the GRE calculator. Take a look at the following problem:

What is the units digit of 3^50?

Before you go scrambling for the calculator, try the following: start small and look for a pattern.

3^1 = 3

3^2 = 9

3^3 = 27

3^4 = 81

3^5 = ??3

3^6 = ??9

3^7 = ???7

First off, don’t let those question marks scare you. Remember, we are only looking for the units digit, so there is no need to know that 3^6 = 729, just that it ends in a 9. So when I go to figure out 3^7, all I have to do is multiply 9 by 3, which is 27, and more importantly, has a units digit of 7.

Now the key to these types of questions is not to multiply all the way up to desired exponent (which would probably eat up the remaining time in the section) but to find a pattern. Notice that the first four numbers, 3, 9, 7, and 1, repeat as soon as you get to 3^5. In other words, every exponent that is a multiple of 4 will end in a 1, just as 3^4 ends in a 1. Every number that is one less than an exponent that is a multiple of 4, will end in a 7, just as 3^3 ends in a 7.

Likewise, every even numbered exponent that is not a multiple of 4 will end in 9, just as 3^2 ends in a 9. Therefore, 3^50 will have a units digit of 9.

Now, let’s take the difficulty level up a notch. Indeed, that is mostly what the GRE quant section does: take familiar concepts and wrap them up in different—and often diabolical—ways.

The following question is very tough, but not tougher than something you might see test day. Take a stab at it!

In the equation, X4^n - Y7^n = p, n is a positive integer, and X and Y can be any integer between 1 and 9, meaning that X4 and Y7 are both two-digit integers. What are all the possible values of the units digit of p if p > 0?

[A] 1

[B] 2

[C] 3

[D] 4

[E] 5

[F] 6

[G] 7

[H] 8

[I] 9


The good news is we don’t have to worry at all about the value of X or Y. Remember the (?) above? The value of the tens digit is irrelevant to the units digit, e.g., 14^2, 74^2, both end in 6.

Therefore, if the value of n is odd, then X4 will always end in 4 (the pattern is 4^1 = 4, 4^2 = 16, 4^3 = 64…notice how the units digits oscillates between 4 and 6). If n is odd, then the units digit of Y7 will be either 7 or 3. Taking the units digit of 4, from X4 when n is odd, we get 4-7 and 4-3, which will give me 7 and 1.

If n is even, then X4 will always end in 6, and Y7 will end in either 9 or 1 (6 – 9 = 7 and 6 – 1 = 5). Therefore, the possible units digits are 1, 5, and 7, or [A], [E], and [G].


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7 Responses to GRE Math – Exponent Challenge

  1. Nick June 3, 2014 at 3:06 pm #

    I’m not understanding the following: “every even numbered exponent that is not a multiple of 4 will end in 9, just as 3^2 ends in a 9. Therefore, 3^50 will have a units digit of 9.”

    50 is also a multiple of 5, and the pattern repeats every fifth spot, so I assumed the units digit will end with 3 since 3^5 ended in 3. Why is that an incorrect way to look at it?

    I did the math and understand that I was totally wrong, I just need clarification on how to approach these types of problems.

  2. Ana December 1, 2013 at 3:20 pm #

    I don’t know if this is a happy coincidence, but the first time I did this problem I did X4^n+Y7^n and I got the same answer. Is there a mathematical reason it’d be symmetrical like that?

    • Chris Lele
      Chris Lele December 2, 2013 at 11:57 am #

      So the different units digits you can get are for X4^n are 4 and 6. And for Y7^n 1, 3, 7, 9.

      4 matches with 7 and 3
      6 matches with 9 and 1

      4+7 and 4+3 have a 1 and 7 respectively for units digits and 4-7 and 4-3 have 7 and 1 for units digits
      6+9 and 6+1 have 5 and 1 and 6-9 and 6-1 have 5 and 5.

      So not quite symmetry but there are really only 4 different units digit we get (1, 5, and 7, and they happen to apply to both addition and subtraction). This is the type of question that might have to be changed to cut muster as an actual question because a question shouldn’t award those who misinterpret the question.

      Of course this question is just for practice!

  3. Hrag September 18, 2013 at 1:50 am #

    In the last question, it is true that when the exponents are odd, X4 will always end in 4, and Y7 will either end in 7 or 3. So according to X4 – Y7, we get 4-7= 3 (not 7 as explained) as the units digit OR 4-3=1.

    Hence, the answers are 1, 3, 5 and 7, right?

    • Chris Lele
      Chris Lele September 20, 2013 at 9:59 am #

      So here’s the little twist — and I guess I should have done a better job of pointing it out in the explanation. If you subtract 7 from the 4, you get 7, not 3. For example, 34-17 = 17, or 54-47 = 7.

      Hope that makes sense!

      • David L. September 27, 2013 at 1:31 am #

        I think Hrag brings up a good point, though.

        The second problem does not state that p > 0 or that x > y so we must consider cases when p < 0, which means the units digit of p can also be 3 or 9.

        For example:
        14 – 17 = -3
        14^3 – 17^3 = -2169

        Unless, for some reason, the GRE doesn't acknowledge the existence of negative numbers. =)

        • Chris Lele
          Chris Lele September 27, 2013 at 11:32 am #

          Hmm…that is a good point. Yes, I missed that–p being negative :). Good catch! I’ll change the problem so that it reads p > 0.

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