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)

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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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

I love this concept

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