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GMAT Quant: Finding the Units Digits of Large Powers

See how you do with this question:

What is the units digit of  57^45?

A) 1

B) 3

C) 5

D) 7

E) 9

 

To solve this, we’ll begin examining smaller powers and look for a pattern.

57^1 = 57 (the units digit is 7)

57^2 = 3249 (the units digit is 9)

57^3 = 185,193  (the units digit is 3)

 

Aside: Since these powers increase quickly, it’s useful to notice that we need only multiply the units digit each time. For example, the units digit of  57^2 is the same as the units digit of  7^2. Similarly, the units digit of  57^3 is the same as the units digit of  7^3.

 

So, once we know that the units of  57^2 is 9, we can find the units digit of  57^3 by multiplying 9 by 7 to get 63. So the units digit of  57^3 is 3.

To find the units digit of  57^4, we’ll multiply 3 by 7 to get 21. So the units digit of  57^4 is 1.

When we start listing the various powers, we can see a pattern emerge:

The units digit of  57^1 is 7

The units digit of  57^2 is 9

The units digit of  57^3 is 3

The units digit of  57^4 is 1

The units digit of  57^5 is 7

At this point, we should recognize that the pattern begins to repeat. The pattern goes: 7-9-3-1-7-9-3-1-7-9-3-1-…

Since the pattern repeats itself every 4 powers, we say that the “cycle” equals 4

Now comes an important observation:

The units digit of  57^1 is 7

The units digit of  57^2 is 9

The units digit of  57^3 is 3

The units digit of  57^4 is 1

The units digit of  57^5 is 7

The units digit of  57^6 is 9

The units digit of  57^7 is 3

The units digit of  57^8 is 1

The units digit of  57^9 is 7

The units digit of  57^10 is 9

The units digit of  57^11 is 3

The units digit of  57^12 is 1. . . etc.

As you can see, since the cycle = 4, the units digit of  57^k will be 1 whenever k is a multiple of 4.

Now to find the units digit of  57^45, all we need to do is recognize that the units digit of  57^44 is 1 (since 44 is a multiple of 4).

From here, we’ll just continue with our pattern:

The units digit of  57^44 is 1

The units digit of  57^45 is 7

The units digit of  57^46 is 9

The units digit of  57^47 is 3 . . . etc.

So, the units digit of  57^45 is 7, which means the answer is D.

 

If you’d like to practice, you can answer these two questions:

  1. What is the units digit of  83^75?
  2. What is the units digit of  39^61?

(The answers can be found at the very bottom of this post)

 

 

 

 

Answers:

1. 7

2. 9

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12 Responses to GMAT Quant: Finding the Units Digits of Large Powers

  1. Pk June 1, 2015 at 2:06 am #

    Nice way of explaining it. Thanks a lot!.. :)

    • Rita Kreig
      Rita Kreig June 2, 2015 at 2:47 pm #

      You’re so welcome! :)

  2. Sirisha January 13, 2015 at 4:08 am #

    I love this concept :-)

  3. Akhil November 5, 2014 at 10:40 pm #

    Thanks it’s very helpful

    • Mike McGarry
      Mike November 10, 2014 at 11:28 am #

      Dear Akhil,
      You are quite welcome! :-) Best of luck to you!
      Mike :-)

  4. kohila October 18, 2012 at 7:31 am #

    your explanation is nice.. but i have a doubt that how to find the unit digit of 9^26. the possibilities of 9 is 9 and 1. so two possibities . 26/2 gives no reminder.. then how to find the unit digit..

    • Mike McGarry
      Mike October 18, 2012 at 1:19 pm #

      Dear Kohila,
      I’m happy to help. :-) Think about the pattern
      9^1 — unit’s digit of 9
      9^2 — unit’s digit of 1
      9^3 — unit’s digit of 9
      9^4 — unit’s digit of 1
      Thus, *odd* powers of 9 have a units digit of 9, and *even* powers of 9 have a units digit of 1. We know 26 is even, so 9^26 has a units digit of 1.
      Does this make sense?
      Mike :-)

  5. kamran April 19, 2012 at 9:25 am #

    an easier way to answer this is..
    once you find the cycle (in the original question the cycle is 4 from 7,9,3,1) then find the remainder of exponent divided by the cycle –> 45/4 –> remainder=1. the final answer can then be calculated by taking the units digit of the following calculation –> units digit in the original question^remainder. in this case, 7^1=7.units digit of 7 is 7. therefore answer is 7.

    using same procedure for 83^75, cycle for units digit 3 is 4 (from 3,9,7,1). then find the remainder of exponent divided by the cycle–>75/4–>remainder=3. take units digit of following calculation–>units digit in the original question^remainder=3^3=27. units digit of 27 is 7, therefore answer is 7.

    for 39^61, cycle for units digit 9 is 2 (from 9,1). then find the remainder of exponent divided by the cycle–> 61/2–>remainder=1. take units digit of following calculation
    –>units digit in the original question^remainder=9^1=9. units digit of 9 is 9. therefore answer is 9.

    • Chris Lele
      Chris April 19, 2012 at 1:46 pm #

      Hi Kamran,

      I think your way is very efficient. In this post, I think Brent was trying to explicate the principle by writing out all the powers of 57. You wouldn’t necessarily do this when solving the problem; you would follow an approach similar to yours.

      So thanks for sharing :).

    • weirdo2989 September 30, 2012 at 8:07 am #

      This is a great info. Thanks a tonne :)

      • Brent Hanneson October 18, 2012 at 7:36 am #

        Thanks for the feedback!

        Cheers,
        Brent

    • Brent Hanneson October 18, 2012 at 7:36 am #

      Hi Kamran,

      Sorry for not responding earlier.
      I’ll echo what Chris said: the instruction was intended to help you find the pattern. As you suggest, we need not write out everything.

      Cheers,
      Brent


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