Find all School-related info fast with the new School-Specific MBA Forum

It is currently 28 May 2016, 23:58
GMAT Club Tests

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Divisibility by 3, 9, 11, 13, 99, 101, 999, 1001.. 10^k +- 1

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

1 KUDOS received
VP
VP
avatar
Joined: 23 Mar 2011
Posts: 1111
Concentration: Healthcare, Strategy
Schools: Duke '16 (M)
Followers: 77

Kudos [?]: 479 [1] , given: 461

Premium Member CAT Tests
Divisibility by 3, 9, 11, 13, 99, 101, 999, 1001.. 10^k +- 1 [#permalink]

Show Tags

New post 15 May 2013, 11:23
1
This post received
KUDOS
Couldn't find this with basic search so posting it.

Basic concepts used:
1.) Abcd = 10^3 x A + 10^2 x b + 10^1 x c + 10^0 x d
2.) (A x B) mod p = (A mod p) X (B mod p) e.g. (13 x 11) mod 4 = (13 mod 4) x (11 mod 4) = 1 x 3 = 3 (or 1 x -1= -1 [=3] )
3.) (A+B) mod p = (A mod p) + (B mod p) e.g. (13 + 11) mod 4 = 13 mod 4 + 11 mod 4 = 1+(-1) =0 (or 1+3 =4 [=0])

Divisibility for 3 and 9:
Adding all the digits and checking if the sum is divisible by 3 or 9 e.g. 234 mod 9 = (2+3+4) mod 9 = 0 hence divisible.

Proof:
Let the number be of the form ‘abcd’.
Abcd mod 9 = (10^3 x A + 10^2 x b + 10^1 x c + 10^0 x d) mod 9
= (10^3) mod 9 x A mod 9 + 10^2 mod 9 x B mod 9 + 10 mod 9 x c mod 9+ d mod 9
= (10 mod 9 )^3 x A mod 9 + (10 mod 9)^2 x B mod 9 + (10 mod 9) x c mod 9 + d mod 9
=1 x A mod 9+ 1 x B mod 9 + 1 x C mod 9 + d mod 9
= (A+b+c+d) mod 9

We were able to do this because we know 10 mod 9 = 1. Similarly if we know the relation between 10^k and m (where k and m are positive numbers) we can find the divisibility rule for (any number) mod m.

For 7:

1000 mod 7 = -1
Let any 9 digit number be: abcdefghi
Abcdefghi= 10^8 x A + 10^7 x B+ …… + 10^0 X i
= 10^6(10^2 x A + 10^1 x B + 10^0 x C) + 10^3 (10^2 x D + 10^1 x E + 10^0 x F) + 10^0 x (10^2 x G + 10^1 x H + 10^0 x I)

Now (abcdefghi) mod 7 =
[10^6(10^2 x A + 10^1 x B + 10^0 x C) + 10^3 (10^2 x D + 10^1 x E + 10^0 x F) + 10^0 x (10^2 x G + 10^1 x H + 10^0 x I)] mod 7

= {(10^6) mod 7} x {(10^2 x A + 10^1 x B + 10^0 x C) (mod 7)} + {10^3 mod 7} x {(10^2 x D + 10^1 x E + 10^0 x F) mod 7} + {10^0 mod 7} x {(10^2 x G + 10^1 x H + 10^0 x I) mod 7 }

Now,
10^3 mod 7 = -1
10^6 mod 7 = (10^3 mod 7) x (10^3 mod 7) = -1 x -1 = 1
10^9 mod 7 = 10^6 mod 7 x 10^3 mod 7 = 1 x -1 = -1

Therefore,
(abcdefghi) mod 7 = (10^2 x A + 10^1 x B + 10^0 x C) mod 7 + (-1) x (10^2 x D + 10^1 x E + 10^0 x F) mod 7 + (10^2 x G + 10^1 x H + 10^0 x I) mod 7

In short, see if (ABC-DEF+GHI) is divisible by 7 or not

Similar for 13 and 1001 because 1000 mod 13 = -1 and 1000 mod 1001 = -1

Similarly, for any number that is of the form 10^k +/- 1. If the number is abcdef:

For 9 : f+e+d+c+b+a because 10 mod 9 =1
For 11: f-e+d-c+b-a or (f+d+b) – (e+c+a) ; because 10 mod 11 = -1
For 99: (ef + cd + ab) because 100 mod 99 =1 e.g. 2345 mod 99 = (23 + 45) mod 99 = 68
For 101: (ef – cd+ ab) because 100 mod 101 = -1 e.g. 2345 mod 101 = (45-23) mod 101 = 22

you may do this with any number of the form (a0a1a2a3...ai) base 10. Basically any number in decimal system. For other systems, similar concept can be used to find the remainder - will be a bit more complex though.


I see aliens...
Senior Manager
Senior Manager
User avatar
Joined: 29 Mar 2012
Posts: 287
Concentration: Entrepreneurship
Schools: Kellogg, Kellogg '19
GMAT 1: 640 Q50 V26
GMAT 2: 660 Q50 V28
GMAT 3: 730 Q50 V38
Followers: 25

Kudos [?]: 322 [0], given: 23

GMAT ToolKit User
Re: Divisibility by 3, 9, 11, 13, 99, 101, 999, 1001.. 10^k +- 1 [#permalink]

Show Tags

New post 15 May 2013, 23:41
Hi,

Simplified (in terms of language)-

Divisibility rule for 3


\(\frac {100a + 10b + c}3\)
100a+10b+c=99a+a+9b+b+c=(99a+9b)+(a+b+c)
(99a+9b) is divisible by 3, so, the number is divisible by 3 when a+b+c is divisible by 3 and that's sum of all digit in a number.

Regards,

Last edited by cyberjadugar on 16 May 2013, 04:47, edited 1 time in total.
VP
VP
avatar
Joined: 23 Mar 2011
Posts: 1111
Concentration: Healthcare, Strategy
Schools: Duke '16 (M)
Followers: 77

Kudos [?]: 479 [0], given: 461

Premium Member CAT Tests
Re: Divisibility by 3, 9, 11, 13, 99, 101, 999, 1001.. 10^k +- 1 [#permalink]

Show Tags

New post 16 May 2013, 03:04
cyberjadugar wrote:
Hi,

Simplified -

Divisibility rule for 3


\(\frac {100a + 10b + c}3\)
100a+10b+c=99a+a+9b+b+c=(99a+9b)+(a+b+c)
(99a+9b) is divisible by 3, so, the number is divisible by 3 when a+b+c is divisible by 3 and that's sum of all digit in a number.

Regards,


Well, this is does not 'simplify' anything. If you look carefully, it is pretty much the same as what I wrote above.

Anyhow, the motive was not to present the simplest approach, but to provide a generic approach.
Re: Divisibility by 3, 9, 11, 13, 99, 101, 999, 1001.. 10^k +- 1   [#permalink] 16 May 2013, 03:04
    Similar topics Author Replies Last post
Similar
Topics:
4 Experts publish their posts in the topic Divisibility by 11 sakchy 7 02 Aug 2015, 11:43
2 Experts publish their posts in the topic Time saving Divisibility Rules for numbers 1-9 chdeepak96 6 30 Jul 2015, 23:44
General Divisibility Rule for any Number Ending 1, 3, 7 or 9 souvik83 0 22 Apr 2014, 08:17
Experts publish their posts in the topic sqrt3 + 1/3+sqrt3 - 1/3-sqrt3 Skag55 1 29 Nov 2013, 05:25
Experts publish their posts in the topic Last 1/3 of OG 12 Problem Solving section hard or easy?? biehniac 3 22 Apr 2011, 11:58
Display posts from previous: Sort by

Divisibility by 3, 9, 11, 13, 99, 101, 999, 1001.. 10^k +- 1

  new topic 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 and phpBB SEO

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®.