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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.

\(\frac {100a + 10b + c}3\) 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, 03:47, edited 1 time in total.

\(\frac {100a + 10b + c}3\) 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.

gmatclubot

Re: Divisibility by 3, 9, 11, 13, 99, 101, 999, 1001.. 10^k +- 1
[#permalink]
16 May 2013, 02:04

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Happy 2017! Here is another update, 7 months later. With this pace I might add only one more post before the end of the GSB! However, I promised that...