It is currently 20 Oct 2017, 20:22

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.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

What will be the remainder when 13^7 + 14^7 + 15^7 + 16^7 is

  post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
Expert Post
GMAT Tutor
avatar
B
Joined: 24 Jun 2008
Posts: 1339

Kudos [?]: 1954 [0], given: 6

Re: funny remainder... [#permalink]

Show Tags

New post 12 Aug 2008, 09:55
x2suresh wrote:
IanStewart wrote:
arjtryarjtry wrote:
What will be the remainder when 13^7 + 14^7 + 15^7 + 16^7 is divided by 58?
57
1
30
0
28


If you know modular arithmetic:

13^7 + 14^7 + 15^7 + 16^7 is clearly even --> divisible by 2

If it's also divisible by 29, it's divisible by 58.

Notice
16 ~ -13 mod 29
15 ~ -14 mod 29

(~ means 'congruent to')
13^7 + 14^7 + 15^7 + 16^7 ~ 13^7 + 14^7 + (-14)^7 + (-13)^7 mod 29
~ 0 mod 29

So the expression is divisible by 2 and 29, and thus by 58.


I couldn't understand the highlighted part.

Could you elaborate


I was using modular arithmetic, which isn't needed on the GMAT. I can attempt a brief summary, but this would normally be a chapter in a book. When you learn modular arithmetic, you learn in which arithmetic situations you can replace one number with another number that gives the same remainder. Here, we are dividing by 29; the remainder you get when you divide 16 by 29 is the same as the remainder you get when you divide 45 by 29, or 74 by 29, or -13 by 29 (adding or subtracting multiples of 29 from any number will not affect the remainder you get when you divide by 29). We say that "16 is congruent to 45 mod 29", which means "16 has the same remainder as 45 when divided by 29". This ensures that 16^7 will have the same remainder as 45^7 when you divide either by 29 (a fact not hard to prove). 16^7 will also have the same remainder as (-13)^7 when divided by 29, and that's what I used in the solution quoted above. Indeed, 16^7 has the same remainder as (-13)^7, 15^7 has the same remainder as (-14)^7, and 13^7 + 14^7 + 15^7 + 16^7 must have the same remainder as 13^7 + 14^7 + (-14)^7 + (-13)^7 when divided by 29. Clearly 13^7 + 14^7 + (-14)^7 + (-13)^7 is equal to zero; thus 13^7 + 14^7 + 15^7 + 16^7 is divisible by 29.
_________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

Kudos [?]: 1954 [0], given: 6

Manager
Manager
User avatar
Joined: 30 Jul 2007
Posts: 129

Kudos [?]: 22 [0], given: 0

Re: funny remainder... [#permalink]

Show Tags

New post 12 Aug 2008, 10:37
If a^n + b^n + c^n + d^n is divisible by a + b + c + d as long as n is odd, then in our case, wouldn't the remainder be 0?

Kudos [?]: 22 [0], given: 0

Re: funny remainder...   [#permalink] 12 Aug 2008, 10:37

Go to page   Previous    1   2   [ 22 posts ] 

Display posts from previous: Sort by

What will be the remainder when 13^7 + 14^7 + 15^7 + 16^7 is

  post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.