NEW HERE? LEARN MORE
closeclose
Last visit was: 23 Apr 2024, 12:50 It is currently 23 Apr 2024, 12:50
Decision Tracker
My Rewards
New posts
New comers' posts

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

When 51^25 is divided by 13, the remainder obtained is:

SORT BY:
Date
Kudos
   1   2   3 
Tags:
Find Similar Topics 
Show Tags
Hide Tags
User avatar
B
Gmatguy007
Manager
Manager
Joined: 07 Feb 2024
Posts: 70
Own Kudos [?]: 4 [0]
Given Kudos: 91
Send PM
When 51^25 is divided by 13, the remainder obtained is: [#permalink] New post   11 Mar 2024, 11:24
 
Quote:
 When 51^25 is divided by 13, the remainder obtained is:

A. 12
B. 10
C. 2
D. 1
E. 0

­Why don't we solve it with the method of cyclicity? As in other similar questions (for instance, remainder of 43^86 divided by 13), I took the unit digit of 51 which has infinite cyclicity (remains 1) and I divided it with the 13 and the remainder is 1.­
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92881
Own Kudos [?]: 618579 [0]
Given Kudos: 81562
Send PM
CAT Tests Forum Quiz Mode
Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink] New post   11 Mar 2024, 11:41
Expert Reply
Gmatguy007 wrote:
Quote:
 When 51^25 is divided by 13, the remainder obtained is:

A. 12
B. 10
C. 2
D. 1
E. 0

­Why don't we solve it with the method of cyclicity? As in other similar questions (for instance, remainder of 43^86 divided by 13), I took the unit digit of 51 which has infinite cyclicity (remains 1) and I divided it with the 13 and the remainder is 1.­



I believe your doubt is addressed here:

https://gmatclub.com/forum/when-51-25-i ... l#p1427592­
User avatar
B
Gmatguy007
Manager
Manager
Joined: 07 Feb 2024
Posts: 70
Own Kudos [?]: 4 [0]
Given Kudos: 91
Send PM
Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink] New post   11 Mar 2024, 12:06
Bunuel wrote:
Gmatguy007 wrote:
Quote:
  When 51^25 is divided by 13, the remainder obtained is:

A. 12
B. 10
C. 2
D. 1
E. 0

­Why don't we solve it with the method of cyclicity? As in other similar questions (for instance, remainder of 43^86 divided by 5), I took the unit digit of 51 which has infinite cyclicity (remains 1) and I divided it with the 13 and the remainder is 1.­
 


I believe your doubt is addressed here:

https://gmatclub.com/forum/when-51-25-i ... l#p1427592­

­
I got it, so whenever we are interested only in the last digit we use cyclicity, whereas when we want to find the remainder of x^y when divided by z we use binomial. The only exception where we can use cyclicity for the remainder is when the divisor is either 2,5,10 like in the example I mentioned.

Thanks!
User avatar
B
Gmatguy007
Manager
Manager
Joined: 07 Feb 2024
Posts: 70
Own Kudos [?]: 4 [0]
Given Kudos: 91
Send PM
When 51^25 is divided by 13, the remainder obtained is: [#permalink] New post   10 Apr 2024, 13:00
KarishmaB wrote:
LM wrote:
When 51^25 is divided by 13, the remainder obtained is:

A. 12
B. 10
C. 2
D. 1
E. 0

The method used by Bunuel above is the best way to get to the answer.

­KarishmaB Bunuel GMATCoachBen & others, I also chose (A) but by thinking that the remainder will be the same as if we divided 51 by 13. I tested with an example: remainder of 5/2 is 1 and remainder of 5^4/2 is 1.

Is my reasoning correct?­
User avatar
L
KarishmaB
Tutor
Joined: 16 Oct 2010
Posts: 14816
Own Kudos [?]: 64880 [1]
Given Kudos: 426
Location: Pune, India
Send PM
Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink] New post   11 Apr 2024, 00:24
1
Kudos
Expert Reply
Gmatguy007 wrote:
KarishmaB wrote:
LM wrote:
When 51^25 is divided by 13, the remainder obtained is:

A. 12
B. 10
C. 2
D. 1
E. 0

The method used by Bunuel above is the best way to get to the answer.

­KarishmaB Bunuel GMATCoachBen & others, I also chose (A) but by thinking that the remainder will be the same as if we divided 51 by 13. I tested with an example: remainder of 5/2 is 1 and remainder of 5^4/2 is 1.

Is my reasoning correct?­

­
This is not correct. Division by 2, 5 or 10 is different and exponents of 5 always end with 5.  
The explanation of that is rather long and you can check it out on Sunday (details in my signature below) but this logic doesn't work with all dividends and divisors.

Consider this:
Divide 8 by 3. Remainder 2.
Divide 8^2 by 3. Remainder 1.
Divide 8^3 by 3. Remainder 2. 
Divide 8^4 by 3. Remainder 1.

There is a cyclicity to it but the remainders are not always the same.  
User avatar
B
Gmatguy007
Manager
Manager
Joined: 07 Feb 2024
Posts: 70
Own Kudos [?]: 4 [0]
Given Kudos: 91
Send PM
Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink] New post   11 Apr 2024, 04:02
 
KarishmaB wrote:
Gmatguy007 wrote:
­KarishmaB Bunuel GMATCoachBen & others, I also chose (A) but by thinking that the remainder will be the same as if we divided 51 by 13. I tested with an example: remainder of 5/2 is 1 and remainder of 5^4/2 is 1.

Is my reasoning correct?­

­
This is not correct. Division by 2, 5 or 10 is different and exponents of 5 always end with 5.  
The explanation of that is rather long and you can check it out on Sunday (details in my signature below) but this logic doesn't work with all dividends and divisors.

Consider this:
Divide 8 by 3. Remainder 2.
Divide 8^2 by 3. Remainder 1.
Divide 8^3 by 3. Remainder 2. 
Divide 8^4 by 3. Remainder 1.

There is a cyclicity to it but the remainders are not always the same.  
 

­Okay, so it was just a coincidence. thank you for the assistance KarishmaB
GMAT Club Bot
gmatclubot
Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink]
11 Apr 2024, 04:02
   1   2   3 
Moderators:
Bunuel
Math Expert
92881 posts
yashikaaggarwal
Senior Moderator - Masters Forum
3137 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne
Main navigation
GMAT Resources
Partners
MBA Resources
www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions