Last visit was: 22 Apr 2026, 05:01 It is currently 22 Apr 2026, 05:01
Home
Messages
Tests
Schools
Events
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
User avatar
EgmatQuantExpert
User avatar
e-GMAT Representative
Joined: 04 Jan 2015
Last visit: 02 Apr 2024
Posts: 3,657
Own Kudos:
20,861
 [41]
Given Kudos: 165
Expert
Expert reply
Posts: 3,657
Kudos: 20,861
 [41]
When 8^48 is divided by 3 and 5 respectively, the remainders are R1 Updated on: 21 Nov 2018, 04:44
Originally posted by EgmatQuantExpert on 21 Nov 2018, 04:00.
Last edited by EgmatQuantExpert on 21 Nov 2018, 04:44, edited 1 time in total.
1
Kudos
Add Kudos
40
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

55% (hard)

Question Stats:

64% (02:00) correct 36% (02:07) wrong based on 887 sessions

History

Date
Time
Result
Not Attempted Yet
Interesting Applications of Remainders – Practice question 2

When \(8^{48}\) is divided by 3 and 5 respectively, the remainders are \(R_1\) and \(R_2\). What is the value of \(R_1\) + \(R_2\)?

    A. 1
    B. 2
    C. 4
    D. 6
    E. 8

To solve question 3: Question 3

To read the article: Interesting Applications of Remainders

ShowHide Answer
Official Answer
B
Most Helpful Reply
User avatar
Chethan92
User avatar
Joined: 18 Jul 2018
Last visit: 21 Apr 2022
Posts: 901
Own Kudos:
1,508
 [14]
Given Kudos: 95
Location: India
Concentration: Operations, General Management
Schools: IIMB EPGP'22 (M)
GMAT 1: 590 Q46 V25
GMAT 2: 690 Q49 V34
WE:Engineering (Energy)
Products:
My Reviews
Schools: IIMB EPGP'22 (M)
GMAT 2: 690 Q49 V34
Posts: 901
Kudos: 1,508
 [14]
When 8^48 is divided by 3 and 5 respectively, the remainders are R1 21 Nov 2018, 04:30
3
Kudos
Add Kudos
11
Bookmarks
Bookmark this Post
Numbers 2, 3, 7 and 8 have a cyclicity of 4.
Numbers 4 and 9 have a cyclicity of 2.
Numbers 0, 1, 5 and 6 have a cyclicity of 1.

8 has a cyclicity of 4. When 48 is divided by 4, the remainder is 0.
When the remainder is 0, then the cyclicity of the number should be considered as the remainder.
Hence \(8^{48} = 8^4\).

8, when divided by 3, gives a negative remainder of -1. then \(-1^4 = 1^4\) = 1.
\(R_1 = 1\)

Also \(8^4 = 64*64\).
The remainder when its divided by 5 is 4*4 = 16/5 = 1.
\(R_2 = 1\)

\(R_1+R_2 = 2\).

B is the answer.
General Discussion
User avatar
EgmatQuantExpert
User avatar
e-GMAT Representative
Joined: 04 Jan 2015
Last visit: 02 Apr 2024
Posts: 3,657
Own Kudos:
20,861
 [3]
Given Kudos: 165
Expert
Expert reply
Posts: 3,657
Kudos: 20,861
 [3]
When 8^48 is divided by 3 and 5 respectively, the remainders are R1 Updated on: 31 Mar 2019, 03:01
Originally posted by EgmatQuantExpert on 25 Nov 2018, 17:43.
Last edited by EgmatQuantExpert on 31 Mar 2019, 03:01, edited 1 time in total.
1
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post

Solution



Given:

When \(8^{48}\) is divided by
    • 3, the remainder is \(R_1\)
    • 5, the remainder is \(R_2\)

To find:
    • The value of \(R_1 + R_2\)

Approach and Working:
    • \(R_1\) = \((\frac{{8^{48}}}{3})_R = [\frac{{(9-1)^{48}}}{3}]_R = [\frac{{(-1)^{48}}}{3}]_R = (\frac{1}{3})_R = 1\)
    • \(R_2\) =\([\frac{{(Units\ digit\ of\ 8^{48)}}}{5}]_R\)

Now, units digit cycle of 8 is 8, 4, 2, 6. As \(8^{48}\) can be written as \(8^{4k}\), the units digit is 6.
    • Hence, \(R_2\) = \((\frac{6}{5})_R\) = 1
    • Therefore, \(R_1 + R_2\) = 1 + 1 = 2

Hence, the correct answer is option B.

Answer: B
User avatar
stne
User avatar
Joined: 27 May 2012
Last visit: 21 Apr 2026
Posts: 1,808
Own Kudos:
2,090
Given Kudos: 678
Posts: 1,808
Kudos: 2,090
When 8^48 is divided by 3 and 5 respectively, the remainders are R1 10 Dec 2018, 12:58
Kudos
Add Kudos
Bookmarks
Bookmark this Post
EgmatQuantExpert
Interesting Applications of Remainders – Practice question 2

When \(8^{48}\) is divided by 3 and 5 respectively, the remainders are \(R_1\) and \(R_2\). What is the value of \(R_1\) + \(R_2\)?

    A. 1
    B. 2
    C. 4
    D. 6
    E. 8

To solve question 3: Question 3

To read the article: Interesting Applications of Remainders

Show more

hi chetan2u,
Can you explain the first part of this question in a better way, how \(\frac{8^{48}}{3}\) gives a remainder of 1?

I understand cyclicity of 8 is 4.

I understand \({8^{48}}\) gives a unit digit of 6 , but 6 divided by 3 gives no remainder , so what am I missing here ?
User avatar
Chethan92
User avatar
Joined: 18 Jul 2018
Last visit: 21 Apr 2022
Posts: 901
Own Kudos:
1,508
 [1]
Given Kudos: 95
Location: India
Concentration: Operations, General Management
Schools: IIMB EPGP'22 (M)
GMAT 1: 590 Q46 V25
GMAT 2: 690 Q49 V34
WE:Engineering (Energy)
Products:
My Reviews
Schools: IIMB EPGP'22 (M)
GMAT 2: 690 Q49 V34
Posts: 901
Kudos: 1,508
 [1]
When 8^48 is divided by 3 and 5 respectively, the remainders are R1 10 Dec 2018, 18:19
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hey stne, let me give a try.

9 is completely divisible by 3, giving a remainder of 0. When 8 is divisible by 3, it gives a remainder of 2 or -1.
And (-1)^48 will be 1. Hence the final remainder.

Hope it helps

Posted from my mobile device
User avatar
chetan2u
User avatar
GMAT Expert
Joined: 02 Aug 2009
Last visit: 22 Apr 2026
Posts: 11,229
Own Kudos:
44,988
 [3]
Given Kudos: 335
Status:Math and DI Expert
Location: India
Concentration: Human Resources, General Management
GMAT Focus 1: 735 Q90 V89 DI81
Products:
My Reviews
Expert
Expert reply
GMAT Focus 1: 735 Q90 V89 DI81
Posts: 11,229
Kudos: 44,988
 [3]
When 8^48 is divided by 3 and 5 respectively, the remainders are R1 10 Dec 2018, 18:34
2
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
stne
EgmatQuantExpert
Interesting Applications of Remainders – Practice question 2

When \(8^{48}\) is divided by 3 and 5 respectively, the remainders are \(R_1\) and \(R_2\). What is the value of \(R_1\) + \(R_2\)?

    A. 1
    B. 2
    C. 4
    D. 6
    E. 8

To solve question 3: Question 3

To read the article: Interesting Applications of Remainders


hi chetan2u,
Can you explain the first part of this question in a better way, how \(\frac{8^{48}}{3}\) gives a remainder of 1?

I understand cyclicity of 8 is 4.

I understand \({8^{48}}\) gives a unit digit of 6 , but 6 divided by 3 gives no remainder , so what am I missing here ?
Show more

Hi...
We are looking for remainders when \(8^{48}\) is divided by 3 and 5..
1) Remainder when \(8^{48}\) is divided by 5
As you said 8 has cylicity of 4 and here the remainder will depend on the units digit..
So units digit are 8,4,2,6... As 48 is 4*12, so units digit will be same as that of 8^4..
Hence units digit will be 6 and remainder will be 1 since 6 divided by 5 gives us 1 as remainder.
2) Remainder when \(8^{48}\) is divided by 5
Here we will use the property of divisibility by 3 : If the sum of digits of number is divisible by 3, the number itself is divisible by 3.
So 8 will leave 2 as remainder...
8^2 or 64 should leave 2^2 or 4, which is equal to 1, as remainder.. 64=3*21+1
Next 8^3 will give 2^3 or 8, which is same as 2...
So the remainders have a cylicity of 2,1,2,1,2,1....
Odd power will give 2 and even power give 1 as remainder.
Here 48 is power and hence even. Therefore, remainder will be 1..

Of course other ways are to convert them in binomial expansion..
8^(48)=(9-1)^48...
When you expand this, all terms will have 9 except last term..
Expansion : 9^48+9^47*(-1)^1+9^46*(-1)^2....+9^(-1)^47+(-1)^48...
Only (-1)^48 is not divisible by 3, remainder = 0+0+0+0...+0+1=1
But you can always find your answer with out use of binomial expansion as GMAT does not test or rely on binomial expansion.

Combined remainder = 1+1=2...

B
User avatar
stne
User avatar
Joined: 27 May 2012
Last visit: 21 Apr 2026
Posts: 1,808
Own Kudos:
2,090
Given Kudos: 678
Posts: 1,808
Kudos: 2,090
When 8^48 is divided by 3 and 5 respectively, the remainders are R1 Updated on: 11 Dec 2018, 03:12
Originally posted by stne on 11 Dec 2018, 03:00.
Last edited by stne on 11 Dec 2018, 03:12, edited 1 time in total.
Kudos
Add Kudos
Bookmarks
Bookmark this Post
chetan2u
stne
EgmatQuantExpert
Interesting Applications of Remainders – Practice question 2

When \(8^{48}\) is divided by 3 and 5 respectively, the remainders are \(R_1\) and \(R_2\). What is the value of \(R_1\) + \(R_2\)?

    A. 1
    B. 2
    C. 4
    D. 6
    E. 8

To solve question 3: Question 3

To read the article: Interesting Applications of Remainders


hi chetan2u,
Can you explain the first part of this question in a better way, how \(\frac{8^{48}}{3}\) gives a remainder of 1?

I understand cyclicity of 8 is 4.

I understand \({8^{48}}\) gives a unit digit of 6 , but 6 divided by 3 gives no remainder , so what am I missing here ?

Hi...
We are looking for remainders when \(8^{48}\) is divided by 3 and 5..
1) Remainder when \(8^{48}\) is divided by 5
As you said 8 has cylicity of 4 and here the remainder will depend on the units digit..
So units digit are 8,4,2,6... As 48 is 4*12, so units digit will be same as that of 8^4..
Hence units digit will be 6 and remainder will be 1 since 6 divided by 5 gives us 1 as remainder.
2) Remainder when \(8^{48}\) is divided by 5
Here we will use the property of divisibility by 3 : If the sum of digits of number is divisible by 3, the number itself is divisible by 3.
So 8 will leave 2 as remainder...
8^2 or 64 should leave 2^2 or 4, which is equal to 1, as remainder.. 64=3*21+1
Next 8^3 will give 2^3 or 8, which is same as 2...
So the remainders have a cylicity of 2,1,2,1,2,1....
Odd power will give 2 and even power give 1 as remainder.
Here 48 is power and hence even. Therefore, remainder will be 1..

Of course other ways are to convert them in binomial expansion..
8^(48)=(9-1)^48...
When you expand this, all terms will have 9 except last term..
Expansion : 9^48+9^47*(-1)^1+9^46*(-1)^2....+9^(-1)^47+(-1)^48...
Only (-1)^48 is not divisible by 3, remainder = 0+0+0+0...+0+1=1
But you can always find your answer with out use of binomial expansion as GMAT does not test or rely on binomial expansion.

Combined remainder = 1+1=2...

B
Show more


Hi chetan2u,

Thank you, but for the second point , remainder when divided by 3 , why can't we use the unit digit concept?

When finding the remainder while dividing by 5 we use the unit digit concept but don't use this concept when dividing by three, why? How do we decide when to use unit digit concept and when not to use unit digit concept?
User avatar
chetan2u
User avatar
GMAT Expert
Joined: 02 Aug 2009
Last visit: 22 Apr 2026
Posts: 11,229
Own Kudos:
44,988
 [2]
Given Kudos: 335
Status:Math and DI Expert
Location: India
Concentration: Human Resources, General Management
GMAT Focus 1: 735 Q90 V89 DI81
Products:
My Reviews
Expert
Expert reply
GMAT Focus 1: 735 Q90 V89 DI81
Posts: 11,229
Kudos: 44,988
 [2]
When 8^48 is divided by 3 and 5 respectively, the remainders are R1 11 Dec 2018, 03:11
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Hi stne,
The multiples of 5 are 5,10,15,20...
So all numbers ending with 5 or 0 are multiple of 5 and only those that end with 5 and 0 are multiple of 5.
Similarly all ending with 0 are multiple of 10.
All ending with even number are a multiple of 2.
But what about 3... The multiple of 3 are 3,6,9,12,15,18,21,24,27,30..
So 6 is a multiple but 16 is not , 26 is not.
That is why the property of divisibility of 3 is SUM of digits of number should be divisible by 3..
Say 23415 :- 2+3+4+1+5=15 -----1+5=6 and 6 is divisible by 3, so 23415 is divisible by 3..
But say 23:- 2+3=5, 5 not divisible by 3, so 23 is not divisible by 3
User avatar
stne
User avatar
Joined: 27 May 2012
Last visit: 21 Apr 2026
Posts: 1,808
Own Kudos:
2,090
Given Kudos: 678
Posts: 1,808
Kudos: 2,090
When 8^48 is divided by 3 and 5 respectively, the remainders are R1 11 Dec 2018, 03:21
Kudos
Add Kudos
Bookmarks
Bookmark this Post
chetan2u
Hi stne,
The multiples of 5 are 5,10,15,20...
So all numbers ending with 5 or 0 are multiple of 5 and only those that end with 5 and 0 are multiple of 5.
Similarly all ending with 0 are multiple of 10.
All ending with even number are a multiple of 2.
But what about 3... The multiple of 3 are 3,6,9,12,15,18,21,24,27,30..
So 6 is a multiple but 16 is not , 26 is not.
That is why the property of divisibility of 3 is SUM of digits of number should be divisible by 3..
Say 23415 :- 2+3+4+1+5=15 -----1+5=6 and 6 is divisible by 3, so 23415 is divisible by 3..
But say 23:- 2+3=5, 5 not divisible by 3, so 23 is not divisible by 3
Show more

Hi chetan2u,

Thank you, will need to brush up my remainder concept .BTW there is a small typo in point number 2 in the link below:

https://gmatclub.com/forum/when-8-48-is ... l#p2188053

You have 5 instead of 3, you may want to correct it, so that others are not confused. Thank you.
User avatar
PierTotti17
User avatar
Joined: 06 Oct 2018
Last visit: 30 Jun 2020
Posts: 30
Own Kudos:
18
 [4]
Given Kudos: 14
Posts: 30
Kudos: 18
 [4]
When 8^48 is divided by 3 and 5 respectively, the remainders are R1 16 Dec 2018, 08:46
4
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Approach for people not familiar with binomial and/or negative remainders:

8^1 = 8
8^2 = 64
8^3 = 512
8^4 = 4,096
8^5 = 32,768

(you should have 8^3 memorised imo).

now, take each power and divide by 3 to determine cyclicity of remainders (then do the same with 5).

8^1 / 3 = r2
8^2 / 3 = r1
8^3 / 3 = r2

(at this point I stopped, but we could test 8^4 for completeness sake).

8^4 / 3 = r1

Pattern established. Therefore 8^48 yields r1.

Do the same for 5.

8^1 / 5 = r3
8^2 / 5 = r4
8^3 / 5 = r2
8^4 / 5 = r1

(pattern probably ends here - you should know 8 has cyclicity of 4. Nonetheless, if your arithmetic is good, then find 8^5 and divide by 5 to confirm).

8^5 / 5 = r3

Therefore, 8^48 yields remainder 1 when divided by 5.

Hence, R1 = 1, R2 = 1, and R1+R2 = 2

B
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,960
Own Kudos:
1,117
Posts: 38,960
Kudos: 1,117
When 8^48 is divided by 3 and 5 respectively, the remainders are R1 11 Sep 2025, 01:22
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.
Moderators:
Bunuel
Math Expert
109740 posts
Krunaal
Tuck School Moderator
853 posts