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

(the units digit is 7)

(the units digit is 9)

(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 is the same as the units digit of . Similarly, the units digit of is the same as the units digit of .

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

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

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

The units digit of is 7

The units digit of is 9

The units digit of is 3

The units digit of is 1

The units digit of 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 is 7

The units digit of is 9

The units digit of is 3

The units digit of is 1

The units digit of is 7

The units digit of is 9

The units digit of is 3

The units digit of is 1

The units digit of is 7

The units digit of is 9

The units digit of is 3

The units digit of is 1. . .etc.

As you can see, since the cycle = 4, the units digit of will be 1whenever k is a multiple of4.

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

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

The units digit of is 1

The units digit of is 7

The units digit of is 9

The units digit of is 3 . . . etc.

So, the units digit of 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 ?

What is the units digit of ?

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

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.

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.

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?

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.

Magoosh Test Prep ExpertApril 26, 2016 at 2:46 am#

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 🙂

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?

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.

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.

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.

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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

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..

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