Raising to a power is iterated multiplication. Luckily, you can find your units digit with a simple multiplication pattern, even when you’re working with large powers. (For a refresh of the multiplication rules for unit digits, see our post on difficult units digits.)

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} = 3,249 (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:

- What is the units digit of 83
^{75}? - 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*

What about finding the units digit of 13 x 13^13?

Thank you. Being multiplied by 13 is tricky since it seems like we have to get the full amount o 13^13 first, which is almost impossible through the scientific calculator and manual calculation.

Hi Josh,

We can apply the same principles of finding a pattern here! Since there is no exam allowed on the quant section of the GMAT, we have to do this by hand anyway, but I found the pattern after 13^5. Once we find the units digit of 13^13, we just need to multiply that by the units digit of 13 in order to find the answer. First, let’s find the pattern:

13^1=13

13^2=169

13^3=2197

13^4=28561

13^5=371293

I’m starting to see a pattern emerge: 3 9 7 1. If I don’t want to trust this yet, I can do one more in order to be sure:

13^6=4826809

This follows the pattern! So, if we write this pattern out we can find the units digit for the 13th number in this series (13^13):

3 9 7 1 3 9 7 1 3 9 7 1 3

That final 3 is the 13th number, so it is the units digit of 13^13. The easiest way to do this problem would be to recognize that 13 x 13^13 is just 13^14, so we know that the units digit must be the next number in the series (9). But I think you were asking a more general question, so also explain the “long-form” approach to a question like this.

Now, let’s think back to how we learned double-digit multiplication in elementary school. I’ll try to draw it out here (hopefully my formatting doesn’t get messed up when I try this!) , but it’s much easier with a pen and paper! Basically, first we multiply each number by the units digit of the second number (2 in the example below), and then we multiply each number by the tens digit of the second number (2 in the example below), but we must add a 0 to the units digit first! Then, we add the two numbers together to get the right answer:

333

x 12

______

666

+3330

______

3996

What’s important to realize here is that the units digit of the answer is (units digit of first number)*(units digit of second number). This will always be true because of that handy 0 that is a “placeholder” in double digit multiplication.

So, as long as we have the units digit of 13^13 (which we just determined to be 3) and the units digit of the number we are multiplying it by (also 3), then we have our answer: 3*3=9.

Either way we slice it, the answer is 9! This is still a challenging problem and requires quite a bit of multiplication, but it’s still quite possible to answer within the time limits of the GMAT 🙂

How to find 19^19^19 ?

That one’s a bit trickier. But the same principles apply. Here’s how to do the problem step-by-step:

First, focus just on the units digit– the 9.

Next, find the units digit pattern for powers of 9:

9^2 = 8

19^3 = 72

99^4 = 6,56

19^5 = 59,04

9So the pattern is 1-9-1-9-1-9, etc, with a units digit of 1 for even powers, and a units digit of 9 for odd powers.

Right there, we know that 19^19 has a units digit of 9. Thus, 19^19 is really (a large number with a units digit of 9)^19. And since you only need to focus on the units digit, you can treat 19^19 as if it were just 9. So to find the units digit of 19^19^19, just imagine 9^19. As we’ve already discussed, 9^19 has a units digit of 9. So 19^19^19 also has a units digit of 9.

Unit digit of 22^471

and easiest way to find it.

Hi Krishna,

The period for the units digits of 22 is 2, 4, 8, 6.

22^1 = …2

22^2 = …4

22^3 = …8

22^4 = …6

22^5 = …2

22^6 = …4

etc.

So we need to consider multiples of 4 for the period. 471 is either 3 greater than 468 (a multiple of 4) or 1 less than 472 (a multiple of 4) so we can use these to solve. I’ll take 472 as my comparison. If we had 22^472 the units digit would be 6, and 471 will be one before this, so our units digit will be 8.

Thank uh soo much brent

I’m confused on th eon 39^61

I keep getting 1 as the units digit

39^1-9

39^2-1

So just find the closest thing which is 39^60-9

So doesn’t 39^61 have to be 1

Hi Ann,

The period of units digits here is quite simple. We have:

39^1 = …9

39^2 = …1

39^3 = …9

39^4 = …1

So every odd number power has a units digit of 9 and every even number power has a units digit of 1. That means that 39^60 = …1 and 39^61 = …9. I hope that helps! 🙂

can u tell me if we get a remainder 0 what we must do

Hi Krishna 🙂

A number with a units digit of zero raised to any power (except 0) will still have a units digit of 0. Let’s look at the smallest such number, 10, to see how that works.

10^0 = 1

10^1 = 10

10^2 = 100

10^3 = 1000

As you can see, the units digit is always 0 except when for 10^0, which is equal to 1. This pattern is observed for any numbers with a units digit of 0.

I hope this helps 🙂

If you get 0 in remainder you can put that number power as 4 and solve the problem.

For Example: 8^64

64/4 = Remainder 0.

So you can put 8^4 that is 8*8*8*8

64*64 eliminate the tenth digit that is 4*4 = 16

So you will get its unit digit 6.

Hope this witl help 🙂

thks helped alot

You’re welcome 🙂 Happy studying!

Hi Mike,

I am little confused here.Practice question 1. 83^ 75

I am getting an answer of 3 and not 7 because 3^5 = has unit digit 3 and it has a pattern of 3971..3971…3971…so 83^74 = will have unit digit 1 and 83^75 = will have unit digit 3

Am I missing anything which is why i am getting the units digit as 3 and not 7?

Hi Swagata,

Happy to help! 🙂

So let’s visualize the pattern you already mentioned:

3^1 = …3

3^2 = …9

3^3 = …7

3^4 = …1

83^74 will not have a units digit of 3 because 74 is not divisible by 4. 83^74 will actually have a units digit of 9 and that means that 83^75 will have a units digit of 7.

I hope that clarifies! 🙂

Cyclicity of 3 is 3971.. Total 4 digits. Now divide 75 by 4 to get 3 as a remainder. for remainder 3 cycle is 3.9.7. Thus ans is 7

Hi mike

I am confuse about 83^75 and try to understand your explanation but i did’t

i get 3 but we i arrange all the number i get 7, i need quick and easy way to solve it.

thank you

Hi Sultan,

The explanation that we provide above is the most efficient way to approach this problem. First, we take the units digit of the 2-digit number we’re interested in (in this case that’s 3). Then we figure out the pattern for the exponents of 3:

3^1 = …3

3^2 = …9

3^3 = …7

3^4 = …1

3^5 = …3

Notice how there are four different units digits that we can get. Now that we know this, we can work towards finding the units digit of 83^75. To see how many times we repeat the pattern, let’s divide 75/4 = 18.75. This means that we go through the series 3, 9, 7, 1 a total of 18 times. And then, we do not go through an entire cycle. We can figure out how far though the cycle we get in a couple of ways. A very straightforward way is to multiple 18*4 = 72 to know that 83^72 has a units digit of 1. We know this because 1 is the last units digit in the cycle. Then we start over again:

83^72 = …1

83^73 = …3

83^74 = …9

83^75 = …7And that’s our answer 😀

I hope this clears up your doubts!

Well done.. Ste by step.. Calculaation..now Easy to solve any example..!!!! Thanks dude..!!

hi

How to find unit digit of the sum 228^128 + 393^193 + 447^147 + 522^122.

When we solve answer is coming 1.

But in book it is given as 6. How it is possible?

Hi Sam 🙂

To find the units digit of this sum, we must first determine the units digit of the individual elements of the sum. We can figure out the units digit of the four elements (228^128, 393^193, 447^147, and 522^122) by following the method explained in this post. For each element, we can determine the cycle of units digits and

then use that information to figure out the units digit for the given exponents. Let’s look at how to do this for the first element, 228^128.

Since the units digit is 8, we can focus on the units digits of the first few powers of 8:

8^1 –> 8

8^2 –> 4

8^3 –> 2

8^4 –> 6

8^5 –> 8

We can see that the units digit has a cycle of 4. Since 128 is a multiple of 4, then we can say that the units digit of 228^128 is 6. Similarly, 3, 7, and 2 also follow cycles of 4, and we can find the units digits of the corresponding elements using the same method as described above. The resulting units digits are:

393^193 –> 3

447^147 –> 3

522^122 –> 4

Adding all of these units digits together, we get a sum of

6+3+3+4 = 16

As you can see, the sum of the units digits is 6 🙂

Hope this helps!

Hi Mike,

I am little confused here.

practice question 1. 83^ 75

The answer should be 3 and not 7.

Am missing anything which is why i am getting the units digit as 3 and not 7?

Hi Charu,

Let’s start with the units digits of the first few powers of 83:

83^1 = 3

83^2 = 9

83^3 = 7

83^4 = 1

83^5 = 3

So using this period of 4 that repeats, we can know that:

83^72 has a units digit of 1

83^73 has a units digit of 3

83^74 has a units digit of 9

83^75 has a units digit of 7

I’m not sure what you are doing to get 3 instead, but hopefully this helps clarify! 🙂

thank you so much ………………..effective one

🙂 Thank you for the step-by-step tutorial, Brent!

Thank u so much for explaining the topic in a good way…so that now I can solve many more other questions

Nice way of explaining it. Thanks a lot!.. 🙂

You’re so welcome! 🙂

units place for 2787^(14^13)?

cycle of 7 completes at 4 i.e units places repeats after that.

7^1=7

7^2=9

7^3=3

7^4=1

14×13=182=180+2

180 is multiple of 4 so unite’s place of (2787)^180=1

now unite’s place of (2787)^2=9

Ans=1×9=9

It is 14^13 NOT 14*13!

Answer is 9.

As (2787)^14^13= 9^13=9^1=9

I love this concept 🙂

Yes, me too..

Hey, sorry to late reply but but the answer is wrong the result is 1 NOT 9

very helpfull for solving such problems.. update some more shortcuts

Thanks it’s very helpful

Dear Akhil,

You are quite welcome! 🙂 Best of luck to you!

Mike 🙂

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..

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 🙂

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.

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 :).

This is a great info. Thanks a tonne 🙂

Thanks for the feedback!

Cheers,

Brent

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