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GMAT Quant: Difficult Units Digits Questions

Mike MᶜGarry

July 17, 2013

GMAT Arithmetic

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

(D) 7

(E) 9

2) The units digit of is:

(A) 2

(B) 4

(D) 8

(E) 0

3) The units digit of is:

(A) 2

(B) 4

(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.

3^1 = 3

3^2 = 9

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

7*3 = 21, so 3^4 has a units digit of 1

1*3 = 3, so 3^5 has a units digit of 3

3*3 = 9, so 3^6 has a units digit of 9

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

7*3 = 21, so 3^8 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: 3^8 , 3^56 , and 3^444 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 3^446 —–

3^444 has a units digit of 1

3^445 has a units digit of 3

3^446 has a units digit of 9

As it happens, 3^446 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,

253^446 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 253^446 , courtesy of Wolfram Alpha:

253^446 = 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

7^1 has a units digit of 7

7^2 has a units digit of 9 (e.g. 7*7 = 49)

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

7^4 has a units digit of 1 (e.g. 7*3 = 21)

7^5 has a units digit of 7

7^6 has a units digit of 9

7^7 has a units digit of 3

7^8 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

7^12 has a units digit of 1

7^13 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.

7^44 has a units digit of 1

7^45 has a units digit of 7

7^46 has a units digit of 9

7^47 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.

3^1 has a units digit of 3

3^2 has a units digit of 9

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

3^4 has a units digit of 1 (e.g. 3*7 = 21)

3^5 has a units digit of 3

3^6 has a units digit of 9

3^7 has a units digit of 7

3^8 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.

3^44 has a units digit of 1

3^45 has a units digit of 3

3^46 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.

4^1 has a units digit of 4

4^2 has a units digit of 6 (e.g. 4*4 = 16)

4^3 has a units digit of 4 (e.g. 4*6 = 24)

4^4 has a units digit of 6

4^5 has a units digit of 4

4^6 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, 44^91 , 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.

3^1 has a units digit of 3

3^2 has a units digit of 9

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

3^4 has a units digit of 1 (e.g. 3*7 = 21)

3^5 has a units digit of 3

3^6 has a units digit of 9

3^7 has a units digit of 7

3^8 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.

3^36 has a units digit of 1

3^37 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, 73^37 , 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

Author

Mike MᶜGarry

Mike served as a GMAT Expert at Magoosh, helping create hundreds of lesson videos and practice questions to help guide GMAT students to success. He was also featured as “member of the month” for over two years at GMAT Club. Mike holds an A.B. in Physics (graduating magna cum laude) and an M.T.S. in Religions of the World, both from Harvard. Beyond standardized testing, Mike has over 20 years of both private and public high school teaching experience specializing in math and physics. In his free time, Mike likes smashing foosballs into orbit, and despite having no obvious cranial deficiency, he insists on rooting for the NY Mets. Learn more about the GMAT through Mike’s Youtube video explanations and resources like What is a Good GMAT Score? and the GMAT Diagnostic Test.

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38 responses to “GMAT Quant: Difficult Units Digits Questions”

Kamrun Nahar

November 25, 2021

Thank you very much Mike for this extremely useful post. After reading this, the units digit problem is clearing up.

Reply
Nitin

July 6, 2019

Thank you for the wonderful explanation!! Really appreciate it!

Reply
Subhashree Ghosh

March 4, 2019

Sir i have a problem in the first question…. My teacher told me that when we have such questions of power on power values at that time we have to first see the 2nd power, if the power is in even number write 2 instead of that power and if the number is odd then write 1 in that place…. So in this question the 2nd power is odd so I write 1 instead of 13 and on the base 137 it’s power is 13…so firstly I divide 13 with 4 and write the reminder on the base 137 and the remainder is 1…so my unit digit that comes is 7 power 1 ie., 7….so how the answer is 3….

Reply
1. Magoosh Test Prep Expert
  
  March 6, 2019
  
  I’ll be honest– I think the approach you’re describing– turning odd powers into 2 and even powers into 1– simply doesn’t work with most base numbers. For example, 2^15 = 32,768, but 2^1 = 2. And 3^11 = 177,147, but 3^1 = 3. For that matter, 7^4 = 2401, but 7^2 = 49. You simply can’t predict the units digit number using the method you’ve outlined. I hope this helps, but let me know if I’ve missed or misunderstood something you said above.
  
  Reply
Nitin

August 5, 2018

This is one of the best explanation for units digit arithmetic questions! Totally loving the learning experience!

Reply
1. Magoosh Test Prep Expert
  
  August 8, 2018
  
  Glad you like it, Nitin! Thanks for sharing your experience 🙂
  
  Reply
Justin

October 14, 2016

For question number 1, why did you not multiply th exponents to equal 611 and then find the units digit for that? I also don’t understand the point of finding the units digit for the inner parentheses when you could just skip to the units digit of the exponent outside the parentheses.

Reply
1. Magoosh Test Prep Expert
  
  October 15, 2016
  
  Hi Justin,
  
  You can definitely approach the question using your approach! The approach above is so you don’t need to work out the multiplication “13” and “47”. However, in some cases, it’s better to just perform the multiplication. To better illustrate the process, we worked out both levels. 🙂
  
  Reply
Ramy

April 9, 2016

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.

Reply
1. Magoosh Test Prep Expert
  
  April 10, 2016
  
  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! 😀
  
  Reply
charlie

March 26, 2016

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?

Reply
1. Magoosh Test Prep Expert
  
  March 27, 2016
  
  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. 🙂
  
  Reply
Varun

March 1, 2016

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

Reply
1. Magoosh Test Prep Expert
  
  March 4, 2016
  
  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” 🙂
  
  Reply
Shikhar

November 16, 2015

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

Reply
Linh

August 10, 2014

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?

Reply
1. Mike
  
  August 11, 2014
  
  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 🙂
  
  Reply
  1. Linh
    
    August 18, 2014
    
    Thank you very much! Your information saves me a lot of time.
    
    Reply
    1. Mike
      
      August 18, 2014
      
      Dear Linh,
      You are quite welcome. 🙂 Best of luck to you!
      Mike 🙂
      
      Reply
2. Elizabeth
  
  April 6, 2019
  
  Thankyou! I learnt it well. The different examples along with their step-by-step solutions really made it simple to understand. I enjoyed solving them🤗
  
  Reply
Edy

March 20, 2014

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.

Reply
1. Mike
  
  March 21, 2014
  
  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
  
  Reply
  1. Saurabh Mittal
    
    August 27, 2015
    
    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.!!
    
    Reply
2. Alisa
  
  August 12, 2014
  
  Thanks, Edy! So much easier =)
  
  Reply
3. Elizabeth
  
  April 6, 2019
  
  Thankyou for this🙆 It saves a lot of time
  
  Reply
Peter

September 27, 2013

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

Reply
1. Mike
  
  September 28, 2013
  
  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 🙂
  
  Reply
Jaydeep

September 9, 2013

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

Reply
1. Mike
  
  September 10, 2013
  
  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 🙂
  
  Reply
Sagnik

August 18, 2013

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?

Reply
1. Mike
  
  August 18, 2013
  
  Dear Sagnik,
  Yes, absolutely. That was a misprint on my part. I just corrected it —- thank you for pointing that out!
  Mike 🙂
  
  Reply
GMATLover

July 20, 2013

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

Reply
1. Mike
  
  July 20, 2013
  
  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 🙂
  
  Reply
  1. Shohidur
    
    November 29, 2013
    
    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
    
    Reply
    1. Mike
      
      November 30, 2013
      
      Dear Shohidur,
      You are more than welcome, my friend. I am very glad you found all this helpful. Best of luck to you.
      Mike 🙂
      
      Reply
  2. mod323
    
    August 23, 2015
    
    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
    
    Reply
alexander

July 18, 2013

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.

Reply
1. Mike
  
  July 18, 2013
  
  Alexander,
  You are more than welcome. Best of luck to you.
  Mike 🙂
  
  Reply

GMAT Quant: Difficult Units Digits Questions

Not what it seems

Units digit arithmetic

Units digits and powers

Summary

Practice problem solutions

Author

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38 responses to “GMAT Quant: Difficult Units Digits Questions”

GMAT Quant: Difficult Units Digits Questions

Not what it seems

Units digit arithmetic

Units digits and powers

Summary

Practice problem solutions

Author

More from Magoosh

Leave a Comment

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38 responses to “GMAT Quant: Difficult Units Digits Questions”