First, a couple 800+ practice questions (yes, you read that right – 800+) on which to whet your chops.

1) The units digit of is:

(A) 1

(B) 3

(C) 5

(D) 7

(E) 9

2) The units digit of is:

(A) 2

(B) 4

(C) 6

(D) 8

(E) 0

3) The units digit of is:

(A) 2

(B) 4

(C) 6

(D) 8

(E) 0

Admittedly, these problems are probably a notch harder than anything you are likely to see on the GMAT. If you understand these, you definitely will understand anything of this variety that the GMAT throws at you!

## Not what it seems

All of those problems above involve numbers with hundreds of decimal places. No one can calculate those answers without a calculator: in fact, no calculator would be sufficient to do the calculation, because no calculator can accommodate that many digits. If one needed the exact answer, one could always use that most extraordinary web computing tool, Wolfram Alpha. Of course, one will not have access to the web or a calculator or anything other than one’s owns wits when confronting a question such as this on the GMAT. How do we proceed?

It turns out, what appears as a ridiculously hard calculation is actually quite easier. No part of the calculation we are going to do will involve anything beyond single-digit arithmetic!

## Units digit arithmetic

The units digits of large numbers are special: they form a kind of elite and exclusive club. The big idea: **only units digits affect units digits**. What do I mean by that? Well, first of all, suppose you add or subtract two large numbers —- **the units digit of the sum or the difference will depend only on the units digits of the two input numbers**. For example, 3 + 5 = 8 —- this means that *any number ending in 3* plus *any number ending in 5* will be a number ending in 8. If you remember your “column addition” processes from grade school, this one might make intuitive sense.

The one that can be a little harder for folks to swallow is multiplication. **If you multiply two large numbers, the unit digit of the product will have the same units digit as the product of the units digit of the two factors**. That’s a mouthful! In other words, let’s take 3*7 = 21, so a units digit of a 3, times a unit digit of a 7, equals a units digit of a 1. That means, we could take any large number ending in 3, times any large number ending in 7, and the product absolutely will have to have a units digit of 1. If this is new idea to you, I strongly recommend: sit down with a calculator and multiply ridiculously large numbers together, with all combinations of units digits, until you are 100% satisfied that this pattern works.

## Units digits and powers

This part will be a recap of an earlier post on powers of units digits. When we raise to a power, of course, that’s iterated multiplication, so we just follow the multiplication rule above. As it turns out, a simple pattern will always emerge.

Suppose we were considering powers of 253 — first of all, only the units digit, 3, matters, for the units digit of the powers. Any number ending in three will have the same sequence of units digits for the powers.

= 3

= 9

9*3 = 27, so has a units digit of 7

7*3 = 21, so has a units digit of 1

1*3 = 3, so has a units digit of 3

3*3 = 9, so has a units digit of 9

9*3 = 27, so has a units digit of 7

7*3 = 21, so has a units digit of 1

Notice a pattern has emerged — 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, … It repeats like mathematical wallpaper. The pattern has a **period** of 4 — that is to say, it take four steps to repeat. This means, 3 to the power of any multiple of 4 has a units digit of 1: , , and all have a units digit of one. If I want to find power that’s not a multiple of 4, that’ easy: I just go to the nearest multiple of 4 and follow the wallpaper from there. For example, if I wanted —–

has a units digit of 1

has a units digit of 3

has a units digit of 9

As it happens, is a number that has 213 digits, but the units digit must be a 9.

The really expansive idea: everything I have just said about powers of 3 also applies to any larger number that happens to have a units digit of 3. Thus,

has a units digit of 9

That number has over a thousand digits (you don’t need to know how to figure that out!) but we know for sure that the units digit of this gargantuan number is 9.

## Summary

If reading this article gave you any insights, you may want to give the questions above a second try. Here’s another problem, slightly easier and more GMAT-like, of this genre:

4) http://gmat.magoosh.com/questions/648

If there is anything you would like to say on this topic, or if you have any questions, please let me know in the comments section.

Finally, on a totally gratuitous note, here, in it’s the full thousand-plus-digit glory, is , courtesy of Wolfram Alpha:

= 6190880832531899190821690500833264378796558621138440416684569816896956548548008694

5501637120599244459832600741641042834529231770919235178215551804306285786976623543

72380800482154741117883167160446214356675850891464689434900627878445424968534812213

04515041521698298553935861735825956938171070649017829532068323699186643091574519875

641384511684642401730622089635116285314952964987090639595147866795944410916642166093

585848058327971863158257619930226042698661632905146850162960633520155118628867911625

239725418415604877699453370534194837316774432534349898272185517986005836675979188507

704257742239368667474408667895362250511057160490511029003928521905584001998500412272

300652930331121107733643816582958394189572596322595033481338694429893546070448926193

272806103607662243076062238492673013489615386273692928543218155895489937520257687664

947027647847750945509362588170852889875925160078794611182855714905968753089053225774

863189760920183769031795458368827168630624310066175033265292467587132663811805301906

7641362643313498166787213628751583911774745199740840719668395714479929

Notice, of course, that the units digit is 9.

## Practice problem solutions

1) First of all, all we need is the last digit of the base, not 137, but just 7. Here’s the power sequence of the units of 7

has a units digit of 7

has a units digit of 9 (e.g. 7*7 = 49)

has a units digit of 3 (e.g. 7*9 = 63)

has a units digit of 1 (e.g. 7*3 = 21)

has a units digit of 7

has a units digit of 9

has a units digit of 3

has a units digit of 1

etc.

The period is 4, so 7 to the power of any multiple of 4 has a units digit of 1

has a units digit of 1

has a units digit of 7

So the inner parenthesis is a number with a units digit of 7.

Now, for the outer exponent, we are following the same pattern — starting with a units digit of 7. The period is still 4.

has a units digit of 1

has a units digit of 7

has a units digit of 9

has a units digit of 3

So the unit digit of the final output is 3. Answer = **B**

BTW, this number is the great-granddaddy, the biggest number of all the big numbers mentioned in this post. The number clocks in with over 1300 digits!

2) We have to figure out each piece separately, and then add them. The first piece is remarkably easy — any power of anything ending in 5 always has a units digit of 5. So the first term has a units digit of 5. Done.

The second term takes a little more work. We can ignore the tens digit, and just treat this base as 3. Here is the units digit patter for the powers of 3.

has a units digit of 3

has a units digit of 9

has a units digit of 7 (e.g. 3*9 = 27)

has a units digit of 1 (e.g. 3*7 = 21)

has a units digit of 3

has a units digit of 9

has a units digit of 7

has a units digit of 1

The period is 4. This means, 3 to the power of any multiple of 4 will have a units digit of 1.

has a units digit of 1

has a units digit of 3

has a units digit of 9

Therefore, the second term has a units digit of 9.

Of course 5 + 9 = 14, so something with a units digit of 5 plus something with a units digit of 9 will have a units digit of 4. Answer = **B**

3) We have to figure out each piece separately, and then multiply them. The powers of 4 are particularly easy.

has a units digit of 4

has a units digit of 6 (e.g. 4*4 = 16)

has a units digit of 4 (e.g. 4*6 = 24)

has a units digit of 6

has a units digit of 4

has a units digit of 6

Four to any odd power will have a units digit of 4. Thus, any number with a units digit of four, raised to an odd power, will also have a units digit of 4. The first factor, , has a units digit of 4.

Now, the base in the second factor ends in a 3 (we can ignore the tens digit). Here is the pattern for powers of three.

has a units digit of 3

has a units digit of 9

has a units digit of 7 (e.g. 3*9 = 27)

has a units digit of 1 (e.g. 3*7 = 21)

has a units digit of 3

has a units digit of 9

has a units digit of 7

has a units digit of 1

The period is 4. This means, 3 to the power of any multiple of 4 will have a units digit of 1.

has a units digit of 1

has a units digit of 3

Thus, any number with a units digit of 7, when raised to the power of 37, will have a units digit of 3. The second factor, , has a units digit of 3.

Of course, 4*3 = 12, so any number with a units digit of 4 times any number with a units digit of 3 will yield a product with a units digit of 2.

Answer = **A**

Hi Mike,

I have a quick question. For the third question, to find out the unit digit for 44 ^ 91..I got 6 as the unit answer. I thought since 90 is a multiple of 5 and 4 ^ 5 has the unit digit of 4 then one more (91) should mean that the unit digit is 6 ??

OR is it only dependent on the repetition of the sequence? Since, it alternates every odd power, that’s what we should use to compare when we look for the final unit digit for 44^91?

I think, I’m just confusing myself at this point..lol BUT it feels good to know that I can at least attempt 800+ questions thanks to your simple and clear explanation.

Hi Ramy,

Happy to help.

In the case you are answering a question like this, the most foolproof way is to find the pattern in the units digits of the powers (aka the period). you tried to use some logic that doesn’t apply to the pattern that you need to answer this question, unfortunately.

If you stick to finding the period and go forward confidently with that, you should have no problem with these types of unit digit questions! 😀

Hello Mr

My name is Charlie. I am a student in Cambodia. I have a problem with an exercise. So the question is find the last six unit digits of 2015^2015? Can you solve it for me please?

Hello Charlie,

Thanks for leaving a comment!

When you say “find the last six unit digits” do you mean the unit digits of the last six numbers leading up to 2015^2015? The units digit is, by definition, only the final digit of any number, so it isn’t possible for a single number to have multiple unit digits.

Let me know and I’ll be happy to help further.

Can you provide the answers to three questions ? Is it B, B, A ?

Yes, you’re correct! So that you know, the answers are also located within the post below the corresponding explanation in the section “Practice problem solutions”

I’m having a hard time dealing with a question,

I found it in class 9 reference book.

The question is related to units digits.

Question : Find the least number having unit digit 7 and if the unit digit is shifted in the beginning of the number (for ex: 123123323’7′ -> ‘7’123123323) it become 5 times the original number.

the way i’m trying :

5(x1x2x3x4x5………x(n)’7′) = ‘7’x1x2x3x4x5…………x(n). [x1, x2, x3, x4, x5…..denotes different digit of a number]

I proceed like:

5*7=35

therefore in place of x(n) 5 will come and (3) will be carried…..

5(x1x2x3x4x5……..5’7′) = ‘7’x1x2x3x4x5……..5

similarly

5*3=15 + (3 carried) = 18

therefore in place of x(n-1) 8 will come and (1) will be carried….

5(x1x2x3x4x5……..85’7′) = ‘7’x1x2x3x4x5……..85

I have tried it till many digit but still i do not find the number specific to the question.

5(91836746938775510204085’7′) = ‘7’91836746938775510204085….and so on.

I would be great if u help me with this question. (i’m solving this question for more than 18hrs.).

Looking forward for Your Kind Help,

Shikhar

Hi Mike,

Instead of finding the last unit digit, do you have any approach to find the last 2 digits? I saw on the GMAT CLUB post by Bunnel:

“Last digits of a product of integers are last digits of the product of last digits of these integers.

For instance last 2 digits of 845*9512*408*613 would be the last 2 digits of 45*12*8*13=540*104=40*4=160=60”

Do you have any faster approach to find the last 2 digits for your examples in this post?

Dear Linh,

I’m happy to respond. My friend, that is 100% not necessary for the GMAT. I have never seen any standardized test ask students to find the last two digits of some complicated product. Everything Bunuel said there is perfectly true mathematically — it’s just that it’s not in the least necessary for the test. There is plenty of math in the world that is well beyond anything that could possibly be on the GMAT. Focus your efforts on what WILL be on the GMAT.

Does all this make sense?

Mike

Thank you very much! Your information saves me a lot of time.

Dear Linh,

You are quite welcome. Best of luck to you!

Mike

no need to find cyclicity!

1- write unit digit of the numbers along with their power.

((137)^13)47)=7^611

2- Always divide the power by 4. (you only need to divide last two digits of power by 4)

11/4 remainder is 3

3- replace the power with the remainder.

7^3=343 unit digit is 3

be careful: if remainder is 0 write power as 4

you can use this method for other questions too.

Edy,

You are obviously a bright individual. What you show here is sophisticated — I’m not sure all students would follow this, but these shortcuts may benefit more advanced students. Thanks for sharing.

Mike

I partially agree with Edy’s solution, one doesn’t need to find the cyclic values every time. But one should always go by dividing the power by 4 and then calculating the values either.

No. 5 and 6 don’t change their unit digits at any power > 0, Numbers 4 and 9 and change their unit digits in a cyclic fashion at every 2nd power.

One should just remember the cyclic turn value of the unit digits which is easy to memorize and use that.

0,1,5,6- Never change unit digit

2,3,7,8- Start repeating unit digits at every 4th power.

4 and 9- Start repeating unit digits at every 2nd power. DONE.!!

Thanks, Edy! So much easier =)

I just had one of those head explosion moments after reading this post.. thanks for the tips!

Peter,

Well, I hope the explosion didn’t make a mess on the floor! I’m very glad you found this helpful. Best of luck to you, my friend.

Mike

Magoosh hi there is much less complicated way of doing this –

35^87 = units digit will be 5

93^46 = we divide 46 by 4(cyclicity of 3 is 4) remainder is 2, 3^2 = 9,9+5=14 , units digit is 4

Dear Deep Jay,

Yes, that is faster —- for those who already are quite comfortable with this topic. Keep in mind, the point of all solutions I show on this blog is not — “Gee, look how fast I can do it because I’m smart!” Rather, they are designed to elucidate all steps so that folks who are still learning so that they can become more comfortable with the topic.

Does this make sense?

Mike

The second factor, 73^37, has a units digit of 2. how is this so? where is the two coming from 73 has a units digit of 3 , no?

Dear Sagnik,

Yes, absolutely. That was a misprint on my part. I just corrected it —- thank you for pointing that out!

Mike

Thanks Mike for the great Explanation. One weird question: does the repetition always occurs after 4 numbers everytime?

Dear GMATLover,

No — four is the longest possible period. Digits {0, 1, 5, 6) always stay the same in powers. The digits {4, 9} has a period of two, simply alternating between two choices in powers. The digits {2, 3, 7, 8) have periods of four, as shown here.

Does this make sense?

Mike

I was reading and felt like an expediency as i did not thank you for the post, But finally i had to say thanks for this part!! amazing, thank you

Dear Shohidur,

You are more than welcome, my friend. I am very glad you found all this helpful. Best of luck to you.

Mike

Hi Mike,

This means I can always assume a cyclicity of 4 which will also cover [0, 1, 5, 6] (actually anything will cover them) and [4, 9] since repeating every 2 also repeat every 4. right?

Thanks

Thanks for sharing your good thoughts with us,your points really helpful for who prepare for GMAT.you provide really wonderful stuff..once again thanks for sharing good information.

Alexander,

You are more than welcome. Best of luck to you.

Mike